Waves and Optics Practice Questions — JAMB UTME — Page 9

Question 161Question →

A beam of monochromatic light travels through a transparent liquid toward a boundary with air. The speed of light in the liquid is $1.80 \times 10^8\text{ m s}^{-1}$ and the speed of light in air is $3.00 \times 10^8\text{ m s}^{-1}$ . What is the value of the sine of the critical angle ( $\sin C$ ) for total internal reflection at this boundary, and under what condition of propagation can total internal reflection occur?

\sin C = 0.60

, and the light ray must travel from the liquid into air.

\sin C = 0.60

, and the light ray must travel from air into the liquid.

\sin C = 1.67

, and the light ray must travel from the liquid into air.

\sin C = 1.67

, and the light ray must travel from air into the liquid.

Show answer & explanation

Answer: $\sin C = 0.60$ , and the light ray must travel from the liquid into air.

Answer

The sine of the critical angle is

\sin C = 0.60

, and total internal reflection can only occur when light travels from the optically denser liquid into air.

Total internal reflection requires light to originate in the optically denser medium (the liquid) and travel toward the rarer medium (air), at an angle of incidence greater than the critical angle. The sine of the critical angle is calculated directly as the ratio of the speed of light in the medium to the speed of light in air:

\sin C = \frac{v_{\text{liquid}}}{v_{\text{air}}} = \frac{1.80 \times 10^8}{3.00 \times 10^8} = 0.60

.

Step-by-Step Solution

1

Determine the refractive index

n

of the liquid relative to air.

n = \frac{c}{v} = \frac{3.00 \times 10^8\text{ m s}^{-1}}{1.80 \times 10^8\text{ m s}^{-1}} = 1.67

(or

\frac{5}{3}

).

Refractive index is the ratio of the speed of light in vacuum/air to the speed of light in the medium.

2

Calculate the sine of the critical angle

\sin C

.

\sin C = \frac{1}{n} = \frac{v}{c} = \frac{1.80 \times 10^8}{3.00 \times 10^8} = 0.60

.

The critical angle relationship between a medium and air is given by

\sin C = \frac{1}{n}

.

3

Identify the necessary physical condition for total internal reflection to take place.

Light must travel from an optically denser medium (liquid) toward an optically less dense medium (air).

For total internal reflection to happen, the ray must bend away from the normal until the angle of refraction reaches

90^\circ

. This only occurs when moving from a denser to a rarer medium.

Key Concept

Critical Angle and Conditions for Total Internal Reflection

Estimated Time:1m 15s

Question 162Question →

A thin diverging lens has a focal length of $20\text{ cm}$ . If it forms an upright image that is one-fourth the size of the object, what is the distance of the object from the lens?

25\text{ cm}

60\text{ cm}

100\text{ cm}

15\text{ cm}

Show answer & explanation

Answer: $60\text{ cm}$

Answer

The distance of the object from the lens is

60\text{ cm}

.

For a diverging (concave) lens, the focal length is negative (

f = -20\text{ cm}

). Since a diverging lens always forms a virtual and upright image, the image distance

v

is negative. Given that linear magnification

m = \frac{|v|}{u} = \frac{1}{4}

, we obtain

v = -\frac{u}{4}

. Substituting these values into the lens formula

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

gives

\frac{1}{-20} = \frac{1}{u} - \frac{4}{u} = -\frac{3}{u}

, which solves to

u = 60\text{ cm}

.

Step-by-Step Solution

1

Identify given parameters and apply optical sign conventions.

Focal length of diverging lens

f = -20\text{ cm}

. Linear magnification

m = \frac{1}{4}

.

Diverging (concave) lenses always have a negative focal length.

2

Express image distance

v

in terms of object distance

u

.

v = -\frac{u}{4}

Magnification

m = \frac{|v|}{u} = \frac{1}{4}

, and since diverging lenses produce virtual images,

v

must be negative.

3

Substitute

f = -20\text{ cm}

and

v = -\frac{u}{4}

into the thin lens formula

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

.

\frac{1}{-20} = \frac{1}{u} + \frac{1}{-\frac{u}{4}} = \frac{1}{u} - \frac{4}{u} = -\frac{3}{u}

Combining terms under a common denominator

u

simplifies the expression.

4

Solve for the object distance

u

.

-\frac{1}{20} = -\frac{3}{u} \implies u = 3 \times 20 = 60\text{ cm}

Cross-multiplying yields the positive object distance of

60\text{ cm}

.

Key Concept

Thin Lens Formula and Optical Sign Conventions for Diverging Lenses

Question 163Question →

A container holds a layer of water of depth $16.0\text{ cm}$ . An immiscible layer of oil of refractive index $1.20$ and thickness $6.0\text{ cm}$ floats on top of the water. If the refractive index of water is $1.33$ (or $\frac{4}{3}$ ), what is the total apparent depth, in centimeters, of a small object resting at the bottom of the container when viewed normally from directly above?

Show answer & explanation

Answer: 17

Answer

The total apparent depth of the object when viewed normally from directly above is

17.0\text{ cm}

.

When an object at the bottom of a container is viewed normally through multiple transparent media, the overall apparent depth is the sum of the apparent depths produced by each medium individually (

d_{\text{app}} = \sum \frac{d_i}{n_i}

). Substituting the given values gives

\frac{16.0}{4/3} + \frac{6.0}{1.20} = 12.0\text{ cm} + 5.0\text{ cm} = 17.0\text{ cm}

.

Step-by-Step Solution

1

Calculate the apparent depth of the water layer.

Apparent depth of water =

12.0\text{ cm}

Apparent depth in a single medium is given by the real depth divided by its refractive index (

d_{\text{app}} = d / n

). For water,

\frac{16.0}{4/3} = 12.0\text{ cm}

.

2

Calculate the apparent depth of the oil layer.

Apparent depth of oil =

5.0\text{ cm}

Using

d_{\text{app}} = d / n

for the oil layer,

\frac{6.0}{1.20} = 5.0\text{ cm}

.

3

Sum the apparent depths of both media.

Total apparent depth =

17.0\text{ cm}

For multiple parallel transparent layers, the total apparent depth is the sum of the apparent depths of each individual layer.

Key Concept

Apparent depth in composite media layers

Question 164Question →

A convex mirror used as a security mirror in a store has a radius of curvature of $20\text{ cm}$ . If a shopper stands $30\text{ cm}$ in front of the mirror, at what distance from the mirror is the shopper's image formed?

7.5\text{ cm}

behind the mirror

15.0\text{ cm}

in front of the mirror

12.0\text{ cm}

behind the mirror

15.0\text{ cm}

behind the mirror

Show answer & explanation

Answer: $7.5\text{ cm}$ behind the mirror

Answer

The shopper's image is formed

7.5\text{ cm}

behind the mirror.

For a convex mirror, the focal length is negative and given by

f = -r/2 = -10\text{ cm}

. Substituting

f = -10\text{ cm}

and object distance

u = +30\text{ cm}

into the mirror equation

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

gives

\frac{1}{v} = -\frac{1}{10} - \frac{1}{30} = -\frac{4}{30}

, which solves to

v = -7.5\text{ cm}

. A negative image distance represents a virtual image formed

7.5\text{ cm}

behind the mirror.

Step-by-Step Solution

1

Determine the focal length of the convex mirror using its radius of curvature

f = -\frac{r}{2} = -\frac{20\text{ cm}}{2} = -10\text{ cm}

By sign convention, a convex mirror has a negative focal length equal to half its radius of curvature.

2

Apply the mirror formula to find the image distance

v

\frac{1}{f} = \frac{1}{u} + \frac{1}{v} \implies \frac{1}{-10} = \frac{1}{30} + \frac{1}{v}

The mirror formula relates focal length (

f

), object distance (

u

), and image distance (

v

).

3

Solve for

\frac{1}{v}

and calculate

v

\frac{1}{v} = -\frac{1}{10} - \frac{1}{30} = -\frac{3 + 1}{30} = -\frac{4}{30} = -\frac{2}{15}\text{ cm}^{-1} \implies v = -7.5\text{ cm}

The negative sign indicates that the image is virtual and located behind the mirror.

Key Concept

Mirror Formula and Sign Convention for Convex Mirrors

Estimated Time:1m 30s

Question 165Question →

A dentist uses a concave mirror to examine a patient's tooth. When the mirror is placed $12\text{ cm}$ in front of the tooth, it forms an erect image that is magnified $3$ times. What is the focal length of the mirror?

18\text{ cm}

9\text{ cm}

24\text{ cm}

36\text{ cm}

Show answer & explanation

Answer: $18\text{ cm}$

Answer

The focal length of the mirror is

18\text{ cm}

.

For a concave mirror, an erect image is always virtual, located behind the mirror. The magnification formula

m = -v/u = +3

gives an image distance of

v = -36\text{ cm}

when the object distance

u = 12\text{ cm}

. Applying the mirror equation

\frac{1}{f} = \frac{1}{u} + \frac{1}{v} = \frac{1}{12} - \frac{1}{36} = \frac{2}{36} = \frac{1}{18}

yields a focal length of

18\text{ cm}

.

Step-by-Step Solution

1

Identify the nature of the image and apply the magnification relationship

Since the concave mirror forms an erect image, the image must be virtual. Therefore, linear magnification

m = -\frac{v}{u} = +3 \implies v = -3u

.

An erect image formed by a spherical mirror is always virtual, which corresponds to a negative image distance under standard sign conventions.

2

Calculate the image distance

v

v = -3 \times 12\text{ cm} = -36\text{ cm}

.

Given object distance

u = +12\text{ cm}

, multiplying by

-3

yields the position of the virtual image behind the mirror.

3

Substitute

u

and

v

into the mirror formula to find

f

\frac{1}{f} = \frac{1}{12} + \frac{1}{-36} = \frac{3 - 1}{36} = \frac{2}{36} = \frac{1}{18} \implies f = +18\text{ cm}.

The mirror equation

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

relates object distance, image distance, and focal length.

Key Concept

Focal length calculation for spherical mirrors producing virtual images

Question 166Question →

A convex spherical mirror always forms a virtual, erect, and diminished image of a real object, regardless of the object's distance from the mirror.

True

False

Show answer & explanation

Answer: True

Answer

The statement is True. A convex mirror consistently produces virtual, erect, and diminished images for all real object positions.

The statement is correct because the outward curvature of a convex mirror causes all incident parallel or diverging rays from a real object to diverge upon reflection. The virtual extensions of these rays converge behind the mirror between the pole and the focus, ensuring the image is always virtual, upright, and smaller than the object.

Step-by-Step Solution

1

Identify the sign conventions for a convex mirror and a real object.

The focal length

f

is negative (

f < 0

) because the focus is behind the mirror, and the object distance

u

is positive (

u > 0

) for a real object.

Establishing proper sign convention is essential for analyzing image formation in curved mirrors.

2

Analyze the mirror equation to determine the sign of the image distance

v

.

From

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

, re-arranging gives

\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = -\left(\frac{1}{|f|} + \frac{1}{u}\right)

. Thus,

v

is always negative.

A negative image distance (

v < 0

) mathematically proves that the image is virtual and located behind the mirror.

3

Evaluate linear magnification

m

to determine image orientation and size.

Using

m = -\frac{v}{u}

, since

v < 0

and

u > 0

,

m > 0

(erect image). Additionally,

|v| = \frac{|f|u}{u + |f|} < u

, so

|m| = \frac{|v|}{u} < 1

(diminished image).

Magnification sign indicates orientation (positive is erect) and magnitude indicates size relative to the object.

Key Concept

Image characteristics in convex mirrors

Question 167Question →

An object placed $30\text{ cm}$ in front of a concave mirror forms a sharp real image on a screen located $60\text{ cm}$ in front of the mirror. If the concave mirror is replaced by a convex mirror having the same magnitude of focal length, where will the image of the object be formed?

12\text{ cm}

behind the mirror

60\text{ cm}

in front of the mirror

60\text{ cm}

behind the mirror

30\text{ cm}

behind the mirror

Show answer & explanation

Answer: $12\text{ cm}$ behind the mirror

Answer

The image is formed

12\text{ cm}

behind the mirror.

First, calculate the focal length of the concave mirror using the mirror equation

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

. Substituting

u = +30\text{ cm}

and

v = +60\text{ cm}

gives

f = 20\text{ cm}

. Next, for a convex mirror of the same focal length magnitude, the focal length is negative (

f = -20\text{ cm}

). Substituting

u = +30\text{ cm}

into the mirror equation gives

\frac{1}{-20} = \frac{1}{30} + \frac{1}{v'}

, which yields

v' = -12\text{ cm}

. The negative sign confirms the image is virtual and located

12\text{ cm}

behind the mirror.

Step-by-Step Solution

1

Determine the focal length of the concave mirror using the mirror formula.

\frac{1}{f} = \frac{1}{u} + \frac{1}{v} = \frac{1}{30} + \frac{1}{60} = \frac{3}{60} \implies f = 20\text{ cm}

.

Both object distance

u = +30\text{ cm}

and real image distance

v = +60\text{ cm}

are positive in front of the mirror.

2

Assign the appropriate sign for the convex mirror's focal length.

f' = -20\text{ cm}

.

Convex mirrors have a virtual focus behind the mirror surface, requiring a negative sign convention.

3

Calculate the image position for the object in front of the convex mirror.

\frac{1}{-20} = \frac{1}{30} + \frac{1}{v'} \implies \frac{1}{v'} = -\frac{1}{20} - \frac{1}{30} = -\frac{5}{60} = -\frac{1}{12} \implies v' = -12\text{ cm}

.

Solving for

v'

yields a negative value, indicating a virtual image formed

12\text{ cm}

behind the convex mirror.

Key Concept

Reflection at spherical mirrors and application of mirror sign conventions

Question 168Question →

A light ray travels near the boundary between medium 1 (refractive index $1.60$ ) and medium 2 (refractive index $1.20$ ). For total internal reflection to occur at this boundary, in which direction must the light travel, and what is the sine of the critical angle ( $\sin C$ )?

From medium 1 to medium 2, with

\sin C = 0.75

From medium 2 to medium 1, with

\sin C = 0.75

From medium 1 to medium 2, with

\sin C = 1.33

From medium 2 to medium 1, with

\sin C = 1.33

Show answer & explanation

Answer: From medium 1 to medium 2, with $\sin C = 0.75$

Answer

Light must travel from medium 1 to medium 2, with

\sin C = 0.75

Total internal reflection can only take place when light travels from an optically denser medium to an optically rarer medium. Here, medium 1 has a higher refractive index (

1.60

) than medium 2 (

1.20

), so light must travel from medium 1 to medium 2. The critical angle relationship

n_1 \sin C = n_2 \sin 90^\circ

yields

\sin C = \frac{n_2}{n_1} = \frac{1.20}{1.60} = 0.75

.

Step-by-Step Solution

1

Determine the required direction of light propagation for total internal reflection.

Light must travel from medium 1 (

n_1 = 1.60

) to medium 2 (

n_2 = 1.20

), which is from an optically denser medium to a rarer medium.

Total internal reflection occurs only when light moves toward a medium of lower refractive index so that the angle of refraction can reach

90^\circ

.

2

Calculate the sine of the critical angle using Snell's law at the critical condition.

\sin C = \frac{n_2}{n_1} = \frac{1.20}{1.60} = 0.75

.

Applying Snell's law

n_1 \sin C = n_2 \sin 90^\circ

gives

\sin C = \frac{n_2}{n_1}

since

\sin 90^\circ = 1

.

Key Concept

Conditions for Total Internal Reflection and Critical Angle Calculation

Estimated Time:1m 0s

Question 169Question →

Two plane mirrors are inclined to each other at an angle of $60^\circ$ . A ray of light strikes the first mirror at an angle of incidence of $40^\circ$ and subsequently undergoes reflection at the second mirror. What is the total angle of deviation, in degrees, produced in the ray after the two successive reflections?

Show answer & explanation

Answer: 240

Answer

The total deviation produced in the light ray after successive reflections from both inclined plane mirrors is

240^\circ

.

When a light ray undergoes successive reflections at two plane mirrors inclined at an angle

\theta

, the total deviation

D

experienced by the ray is given by

D = 360^\circ - 2\theta

. Substituting

\theta = 60^\circ

yields

D = 360^\circ - 2(60^\circ) = 240^\circ

. Step-by-step ray tracing confirms

d_1 = 180^\circ - 2(40^\circ) = 100^\circ

and

d_2 = 180^\circ - 2(20^\circ) = 140^\circ

, giving a total deviation of

100^\circ + 140^\circ = 240^\circ

.

Step-by-Step Solution

1

Calculate the deviation at the first mirror

d_1 = 100^\circ

The angle of deviation for a single reflection at a plane mirror is

d = 180^\circ - 2i

. For

i_1 = 40^\circ

,

d_1 = 180^\circ - 80^\circ = 100^\circ

.

2

Find the angle of incidence on the second mirror using geometry

i_2 = 20^\circ

The glancing angle on the first mirror is

g_1 = 90^\circ - 40^\circ = 50^\circ

. In the triangle formed by the ray path and the two mirrors inclined at

60^\circ

, the glancing angle on the second mirror is

g_2 = 180^\circ - (60^\circ + 50^\circ) = 70^\circ

. Thus, the angle of incidence on the second mirror is

i_2 = 90^\circ - 70^\circ = 20^\circ

.

3

Calculate the deviation at the second mirror and total deviation

d_2 = 140^\circ

,

D = 240^\circ

The deviation at the second mirror is

d_2 = 180^\circ - 2(20^\circ) = 140^\circ

. Since both reflections rotate the ray in the same sense, the total deviation is

D = d_1 + d_2 = 100^\circ + 140^\circ = 240^\circ

.

Key Concept

Total deviation produced by reflection from two inclined plane mirrors is independent of the initial angle of incidence and is given by

D = 360^\circ - 2\theta

, where

\theta

is the angle of inclination between the mirrors.

Question 170Question →

A water wave moving horizontally across the surface of a pond causes a floating cork to move vertically up and down. The cork completes one full oscillation every $0.50\text{ s}$ , and the horizontal distance between two consecutive wave crests is $1.2\text{ m}$ . What is the speed of propagation of the wave, and how is the wave classified?

2.4\text{ m s}^{-1}

, transverse mechanical wave

2.4\text{ m s}^{-1}

, longitudinal mechanical wave

0.60\text{ m s}^{-1}

, transverse mechanical wave

2.4\text{ m s}^{-1}

, transverse electromagnetic wave

Show answer & explanation

Answer: $2.4\text{ m s}^{-1}$ , transverse mechanical wave

Answer

2.4\text{ m s}^{-1}

, transverse mechanical wave

The frequency of oscillation is

f = \frac{1}{T} = \frac{1}{0.50\text{ s}} = 2.0\text{ Hz}

. Using the wave formula

v = f\lambda

, the propagation speed is

v = 2.0\text{ Hz} \times 1.2\text{ m} = 2.4\text{ m s}^{-1}

. Because the cork vibrates vertically while the wave propagates horizontally, the oscillations are perpendicular to the propagation direction, defining a transverse wave. Since water is required for propagation, it is a mechanical wave.

Step-by-Step Solution

1

Determine the frequency of the wave from the period.

The period

T = 0.50\text{ s}

, so frequency

f = \frac{1}{T} = \frac{1}{0.50} = 2.0\text{ Hz}

.

Frequency is the reciprocal of the time period of oscillation.

2

Calculate the wave speed using the wave equation.

v = f \lambda = 2.0\text{ Hz} \times 1.2\text{ m} = 2.4\text{ m s}^{-1}

.

Wave speed equals the product of frequency and wavelength.

3

Classify the wave based on particle motion and medium requirement.

Particle displacement is perpendicular to wave direction (transverse) and requires water as a medium (mechanical).

Perpendicular oscillation defines transverse waves, and medium dependence defines mechanical waves.

Key Concept

Wave speed calculation and wave classification based on vibration direction and medium requirements.

Question 171Question →

A musical note played on an instrument produces a fundamental frequency of $440\text{ Hz}$ . Given that the speed of sound in air is $330\text{ m/s}$ , what is the wavelength, in meters, of the sound wave produced in air corresponding to its second overtone?

Show answer & explanation

Answer: 0.25

Answer

0.25 m

The fundamental frequency (

f_1 = 440\text{ Hz}

) is the first harmonic. The second overtone is the third harmonic, which has a frequency of

3 \times 440\text{ Hz} = 1320\text{ Hz}

. Substituting this into the wave speed equation

\lambda = \frac{v}{f}

gives

\lambda = \frac{330\text{ m/s}}{1320\text{ Hz}} = 0.25\text{ m}

.

Step-by-Step Solution

1

Determine the harmonic number for the second overtone

The second overtone is the third harmonic (

n = 3

)

Overtones are integer harmonics above the fundamental frequency (

n = 1

). Thus, the 1st overtone is

n = 2

and the 2nd overtone is

n = 3

.

2

Calculate the frequency of the second overtone

f_3 = 3 \times 440\text{ Hz} = 1320\text{ Hz}

The frequency of the

n

-th harmonic is

n

times the fundamental frequency.

3

Calculate the wavelength using the wave speed equation

\lambda = \frac{v}{f_3} = \frac{330\text{ m/s}}{1320\text{ Hz}} = 0.25\text{ m}

Wavelength is determined by dividing the speed of sound by the wave frequency.

Key Concept

Harmonics, Overtones, and Wave Equation

Question 172Question →

A surveyor standing $255\text{ m}$ away from the base of a vertical canyon wall emits a short acoustic signal. If the speed of sound in air is $340\text{ m/s}$ , after what time interval will the surveyor detect the reflected echo?

0.75\text{ s}

1.50\text{ s}

1.33\text{ s}

3.00\text{ s}

Show answer & explanation

Answer: $1.50\text{ s}$

Answer

1.50\text{ s}

Sound must travel to the cliff face and reflect back to the surveyor, covering a total distance of

2 \times 255\text{ m} = 510\text{ m}

. Using the speed formula

t = \frac{d}{v}

, the elapsed time is

\frac{510\text{ m}}{340\text{ m/s}} = 1.50\text{ s}

.

Step-by-Step Solution

1

Determine the total distance traveled by the sound wave

s_{total} = 2 \times 255\text{ m} = 510\text{ m}

An echo requires sound to travel to the obstacle and reflect back to the source.

2

Calculate the time taken using the wave speed formula

t = \frac{s_{total}}{v} = \frac{510\text{ m}}{340\text{ m/s}} = 1.50\text{ s}

Time is equal to total distance divided by the speed of propagation.

Key Concept

Echo reflection and two-way sound propagation distance

Estimated Time:1m 0s

Question 173Question →

A sonometer wire of length $0.80\text{ m}$ and mass $2.0\text{ g}$ is maintained under a tension of $100\text{ N}$ . What is the fundamental frequency of vibration of the wire?

125\text{ Hz}

250\text{ Hz}

62.5\text{ Hz}

12.5\text{ Hz}

Show answer & explanation

Answer: $125\text{ Hz}$

Answer

The fundamental frequency of vibration of the wire is

125\text{ Hz}

.

The linear mass density is

\mu = \frac{0.002\text{ kg}}{0.80\text{ m}} = 0.0025\text{ kg/m}

. The velocity of waves on the string is

v = \sqrt{\frac{100}{0.0025}} = 200\text{ m/s}

. The fundamental frequency is

f = \frac{v}{2L} = \frac{200}{2 \times 0.80} = 125\text{ Hz}

.

Step-by-Step Solution

1

Convert mass to kilograms and calculate linear mass density

(\mu)

.

m = 2.0\text{ g} = 0.002\text{ kg}

. Thus,

\mu = \frac{m}{L} = \frac{0.002\text{ kg}}{0.80\text{ m}} = 0.0025\text{ kg/m} = 2.5 \times 10^{-3}\text{ kg/m}

.

Standard SI units must be used for tension in Newtons and length in meters.

2

Calculate the speed of the transverse wave on the string

(v)

.

v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{100}{0.0025}} = \sqrt{40000} = 200\text{ m/s}

.

Wave speed on a stretched string depends directly on tension and inversely on linear density.

3

Calculate the fundamental frequency

(f_1)

.

f_1 = \frac{v}{2L} = \frac{200}{2 \times 0.80} = \frac{200}{1.6} = 125\text{ Hz}

.

For a fixed string vibrating in its fundamental mode, the length equals half the wavelength (

\,L = \frac{\lambda}{2}\,

).

Key Concept

Fundamental frequency of a stretched string

Question 174Question →

A sinusoidal progressive wave propagating through an elastic medium is described by the equation $y = 0.04 \sin(60\pi t - 4\pi x)$ , where $x$ and $y$ are in meters and $t$ is in seconds. As the wave enters a second medium, its wave speed increases by $50\%$ . What is the wavelength of the wave in the second medium?

0.33\text{ m}

0.50\text{ m}

0.75\text{ m}

1.50\text{ m}

Show answer & explanation

Answer: $0.75\text{ m}$

Answer

The wavelength of the wave in the second medium is

0.75\text{ m}

.

Comparing

y = 0.04 \sin(60\pi t - 4\pi x)

to the standard wave equation

y = A \sin(\omega t - k x)

gives

\omega = 60\pi\text{ rad/s}

and

k = 4\pi\text{ rad/m}

. The source frequency is

f = \frac{\omega}{2\pi} = 30\text{ Hz}

, and the initial wavelength is

\lambda_1 = \frac{2\pi}{k} = 0.50\text{ m}

. The initial wave speed is

v_1 = \frac{\omega}{k} = 15\text{ m/s}

. When the wave enters the second medium, its speed increases by

50\%

to

v_2 = 22.5\text{ m/s}

. Frequency is a source property and remains constant (

30\text{ Hz}

) across media boundaries, so the new wavelength becomes

\lambda_2 = \frac{v_2}{f} = \frac{22.5}{30} = 0.75\text{ m}

.

Step-by-Step Solution

1

Extract angular frequency and wave number from the progressive wave equation.

From

y = 0.04 \sin(60\pi t - 4\pi x)

, we find

\omega = 60\pi\text{ rad/s}

and

k = 4\pi\text{ rad/m}

.

Standard form of a harmonic progressive wave is

y = A \sin(\omega t - k x)

.

2

Calculate the frequency and initial wavelength in the first medium.

Frequency

f = \frac{\omega}{2\pi} = \frac{60\pi}{2\pi} = 30\text{ Hz}

. Initial wavelength

\lambda_1 = \frac{2\pi}{k} = \frac{2\pi}{4\pi} = 0.50\text{ m}

.

Wave frequency and wavelength are related to angular parameters by

\omega = 2\pi f

and

k = \frac{2\pi}{\lambda}

.

3

Determine the initial wave speed and the new speed in the second medium.

Initial speed

v_1 = f \lambda_1 = 30 \times 0.50 = 15\text{ m/s}

. New speed

v_2 = 15 \times (1 + 0.50) = 22.5\text{ m/s}

.

Wave speed increases by

50\%

upon entering the second medium.

4

Apply boundary condition for wave refraction to determine the new wavelength.

Since frequency is invariant across media boundaries (

f_2 = f_1 = 30\text{ Hz}

),

\lambda_2 = \frac{v_2}{f_2} = \frac{22.5}{30} = 0.75\text{ m}

.

Frequency depends solely on the wave source and does not change when passing between media.

Key Concept

Frequency invariance and wavelength alteration across media boundaries in progressive waves

Question 175Question →

A transverse progressive wave traveling along a taut string is represented by the equation $y(x, t) = 0.04 \sin(200t - 8x)$ , where $x$ and $y$ are measured in meters and $t$ is in seconds. What is the maximum transverse speed of a particle on the string in meters per second?

Show answer & explanation

Answer: 8

Answer

The maximum transverse speed of a particle on the string is

8.0\text{ m/s}

.

The maximum speed of a particle executing simple harmonic motion as part of a progressive wave is given by

v_{p,\text{max}} = A\omega

. From the wave equation

y(x, t) = 0.04 \sin(200t - 8x)

, the amplitude is

A = 0.04\text{ m}

and the angular frequency is

\omega = 200\text{ rad/s}

. Multiplying these values yields

v_{p,\text{max}} = 0.04 \times 200 = 8.0\text{ m/s}

.

Step-by-Step Solution

1

Compare the given wave displacement equation with the standard progressive wave form.

Standard form:

y(x, t) = A \sin(\omega t - kx)

. Matching coefficients gives

A = 0.04\text{ m}

,

\omega = 200\text{ rad/s}

, and

k = 8\text{ rad/m}

.

Extracting amplitude and angular frequency is essential for evaluating particle motion.

2

Differentiate displacement with respect to time to find the expression for particle velocity.

v_p(x, t) = \frac{\partial y}{\partial t} = A \omega \cos(\omega t - kx)

.

Particle velocity represents the time rate of change of transverse displacement.

3

Determine maximum particle speed by setting the magnitude of the cosine factor to 1.

v_{p,\text{max}} = A \omega = 0.04 \text{ m} \times 200 \text{ rad/s} = 8.0 \text{ m/s}

.

The maximum absolute value of the cosine function is 1.

Key Concept

Maximum particle velocity in a progressive wave (

v_{p,\text{max}} = A\omega

)

Estimated Time:1m 30s

Question 176Question →

A uniform wire of length $0.60\text{ m}$ fixed at both ends vibrates in its fundamental mode with a frequency of $150\text{ Hz}$ . If the length of the wire is reduced to $0.40\text{ m}$ while the tension in the wire is increased by a factor of 4, what is the new fundamental frequency of vibration?

200\text{ Hz}

300\text{ Hz}

450\text{ Hz}

900\text{ Hz}

Show answer & explanation

Answer: $450\text{ Hz}$

Answer

The new fundamental frequency of the vibrating wire is

450\text{ Hz}

.

The fundamental frequency of a stretched string is inversely proportional to its length and directly proportional to the square root of its tension (

f \propto \frac{\sqrt{T}}{L}

). Reducing the length from

0.60\text{ m}

to

0.40\text{ m}

increases frequency by a factor of

\frac{0.60}{0.40} = 1.5

. Quadrupling the tension increases frequency by a factor of

\sqrt{4} = 2

. Combining both effects gives an overall frequency multiplier of

1.5 \times 2 = 3.0

, leading to

3.0 \times 150\text{ Hz} = 450\text{ Hz}

.

Step-by-Step Solution

1

Write the general formula for the fundamental frequency of a stretched string.

f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}

, where

L

is length,

T

is tension, and

\mu

is linear density.

Establishes the functional dependence of frequency on length and tension.

2

Set up the ratio between the new frequency

f_2

and initial frequency

f_1

.

\frac{f_2}{f_1} = \frac{L_1}{L_2} \sqrt{\frac{T_2}{T_1}}

.

Since linear mass density

\mu

remains constant, comparing ratios isolates the changing variables.

3

Substitute the given parameters into the ratio equation.

\frac{f_2}{150} = \frac{0.60}{0.40} \times \sqrt{4} = 1.5 \times 2 = 3.0

.

Calculates the scaling factor for the new fundamental frequency.

4

Solve for the new fundamental frequency

f_2

.

f_2 = 3.0 \times 150 = 450\text{ Hz}

.

Yields the final numerical value of the modified fundamental frequency.

Key Concept

Fundamental frequency of vibrating stretched strings under varying length and tension

Question 177Question →

A beam of monochromatic light of wavelength $6.25 \times 10^{-7}\text{ m}$ is incident normally on a plane diffraction grating having $400\text{ lines/mm}$ . What is the angle of diffraction for the second-order principal maximum?

14.5^\circ

30.0^\circ

45.0^\circ

60.0^\circ

Show answer & explanation

Answer: $30.0^\circ$

Answer

The angle of diffraction for the second-order principal maximum is

30.0^\circ

.

The grating spacing is

d = \frac{1}{400\times 10^3\text{ m}^{-1}} = 2.50 \times 10^{-6}\text{ m}

. Using

d \sin\theta = n\lambda

for

n=2

and

\lambda = 6.25 \times 10^{-7}\text{ m}

gives

\sin\theta = \frac{2 \times 6.25 \times 10^{-7}}{2.50 \times 10^{-6}} = 0.50

, corresponding to

\theta = 30.0^\circ

.

Step-by-Step Solution

1

Calculate the grating spacing (

d

).

d = \frac{1\text{ mm}}{400} = 2.50 \times 10^{-3}\text{ mm} = 2.50 \times 10^{-6}\text{ m}

.

Grating spacing

d

is the reciprocal of the line density

N

per unit length.

2

Apply the diffraction grating equation for the second order (

n=2

).

d \sin\theta = n\lambda \implies (2.50 \times 10^{-6}) \sin\theta = 2 \times (6.25 \times 10^{-7}) = 1.25 \times 10^{-6}

.

The principal maximum condition relates slit spacing, diffraction angle, spectral order, and wavelength.

3

Solve for the diffraction angle

\theta

.

\sin\theta = \frac{1.25 \times 10^{-6}}{2.50 \times 10^{-6}} = 0.50 \implies \theta = \arcsin(0.50) = 30.0^\circ

.

Taking the inverse sine of

0.50

yields the exact angle of diffraction.

Key Concept

Diffraction Grating Equation (

d \sin\theta = n\lambda

)

Question 178Question →

Match each vibrating acoustic system setup on the left with its corresponding displacement standing wave characteristics (number of nodes, antinodes, and wavelength $\lambda$ in terms of pipe or string length $L$ ) on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Pipe closed at fundamental matches 1 node, 1 antinode ($\lambda = 4L$); Pipe open at fundamental matches 1 node, 2 antinodes ($\lambda = 2L$); String fixed at second harmonic matches 3 nodes, 2 antinodes ($\lambda = L$); Pipe closed at first overtone matches 2 nodes, 2 antinodes ($\lambda = \frac{4}{3}L$).

Each standing wave profile is uniquely determined by boundary constraints: fixed ends and closed pipe ends form displacement nodes, whereas open pipe ends form displacement antinodes. Counting the number of quarter-wavelength segments yields the exact relationship between wavelength $\lambda$ and system length $L$.

Step-by-Step Solution

1

Identify boundary conditions for displacement standing waves

Fixed ends of strings and closed ends of pipes are displacement nodes. Open ends of pipes are displacement antinodes.

Physical constraints prevent particle displacement at rigid boundaries while allowing maximum oscillation amplitude at open boundaries.

2

Calculate node/antinode count and wavelength for fundamental modes

For a closed pipe fundamental, $L = \frac{\lambda}{4}$, so $\lambda = 4L$ (1 node, 1 antinode). For an open pipe fundamental, $L = \frac{\lambda}{2}$, so $\lambda = 2L$ (1 node, 2 antinodes).

The distance between a consecutive node and antinode is $\frac{\lambda}{4}$, while the distance between two consecutive antinodes is $\frac{\lambda}{2}$.

3

Calculate node/antinode count and wavelength for higher harmonics

For a string fixed at both ends in the 2nd harmonic, two complete half-wavelength loops exist ($L = \lambda$), giving 3 nodes and 2 antinodes. For a closed pipe in its first overtone (3rd harmonic), $L = \frac{3\lambda}{4}$, so $\lambda = \frac{4}{3}L$, giving 2 nodes and 2 antinodes.

The harmonic number dictates how many quarter-wavelength or half-wavelength segments fit within length $L$.

Key Concept

Boundary conditions, node-antinode distributions, and wavelength formulas for standing waves in strings and pipes

Question 179Question →

A progressive wave traveling through a primary medium is described by the displacement equation $y = 0.05 \sin(80\pi t - 4\pi x)$ , where $x$ and $y$ are in meters and $t$ is in seconds. If the wave enters a secondary medium where its speed decreases to $15\text{ m s}^{-1}$ , what is the wavelength of the wave in the secondary medium?

0.375\text{ m}

0.500\text{ m}

0.188\text{ m}

2.667\text{ m}

Show answer & explanation

Answer: $0.375\text{ m}$

Answer

The wavelength of the wave in the secondary medium is

0.375\text{ m}

.

The option specifying

0.375\text{ m}

is correct because wave frequency

f

remains constant when moving between different media. Extracting

f = \frac{80\pi}{2\pi} = 40\text{ Hz}

from the initial equation allows direct calculation of the new wavelength using

\lambda_2 = \frac{v_2}{f} = \frac{15}{40} = 0.375\text{ m}

.

Step-by-Step Solution

1

Extract angular frequency and wavenumber from the wave equation

From

y = 0.05 \sin(80\pi t - 4\pi x)

, we identify

\omega = 80\pi\text{ rad s}^{-1}

and

k = 4\pi\text{ rad m}^{-1}

.

The standard wave equation form is

y = A \sin(\omega t - k x)

.

2

Determine the constant frequency of the wave

f = \frac{\omega}{2\pi} = \frac{80\pi}{2\pi} = 40\text{ Hz}

.

Frequency depends solely on the wave source and remains unchanged when crossing media boundaries.

3

Calculate the wavelength in the secondary medium using the new speed

\lambda_2 = \frac{v_2}{f} = \frac{15\text{ m s}^{-1}}{40\text{ Hz}} = 0.375\text{ m}

.

The wave speed formula

v = f \lambda

rearranged gives

\lambda = \frac{v}{f}

.

Key Concept

Wave speed changes across media boundaries while frequency remains invariant

Question 180Question →

A bat emitting an ultrasonic sound pulse towards a flat vertical wall receives the reflected echo $0.12\text{ s}$ after emission. If the speed of sound in air is $340\text{ m/s}$ , what is the distance between the bat and the wall at the moment of emission?

20.4\text{ m}

40.8\text{ m}

81.6\text{ m}

10.2\text{ m}

Show answer & explanation

Answer: $20.4\text{ m}$

Answer

The distance between the bat and the wall is

20.4\text{ m}

.

The distance to the wall is

20.4\text{ m}

because sound undergoes a two-way journey (from source to reflector and back). Multiplying the speed of sound (

340\text{ m/s}

) by the total time (

0.12\text{ s}

) yields the round-trip distance of

40.8\text{ m}

. Dividing this value by

2

gives the one-way distance of

20.4\text{ m}

.

Step-by-Step Solution

1

Calculate the total distance traveled by the sound wave during the round trip.

d_{\text{total}} = v \times t = 340\text{ m/s} \times 0.12\text{ s} = 40.8\text{ m}

The sound wave travels from the bat to the obstacle and reflects back to the bat.

2

Divide the total round-trip distance by

2

to determine the one-way distance to the wall.

d = \frac{d_{\text{total}}}{2} = \frac{40.8\text{ m}}{2} = 20.4\text{ m}

An echo involves a two-way path, so the one-way distance is half the total distance covered by the wave.

Key Concept

Echo and Two-Way Distance Calculation

Estimated Time:1m 30s

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