Waves and Optics Practice Questions — JAMB UTME — Page 5

Question 81Question →

A vibrating tuning fork generates a longitudinal sound wave in air. The fork completes $120$ full oscillations in $0.40\text{ s}$ . If the wavelength of the sound wave in air is $1.15\text{ m}$ , what is the speed of propagation of the wave in $\text{m/s}$ ?

Show answer & explanation

Answer: 345

Answer

The speed of propagation of the sound wave in air is

345\text{ m/s}

.

The frequency of oscillation is determined by dividing the number of oscillations by the total time taken:

f = \frac{120}{0.40\text{ s}} = 300\text{ Hz}

. Substituting the frequency and given wavelength into the wave equation

v = f \lambda

yields

v = 300\text{ Hz} \times 1.15\text{ m} = 345\text{ m/s}

.

Step-by-Step Solution

1

Determine the frequency of the longitudinal wave.

f = 300\text{ Hz}

Frequency is the number of complete oscillations per unit time:

f = \frac{N}{t} = \frac{120}{0.40\text{ s}} = 300\text{ Hz}

.

2

Calculate the wave propagation speed using the wave equation.

v = 345\text{ m/s}

The speed of a progressive wave is given by

v = f \lambda = 300\text{ Hz} \times 1.15\text{ m} = 345\text{ m/s}

.

Key Concept

Relationship between frequency, wavelength, and wave propagation speed in a mechanical medium

Question 82Question →

A wave traveling through a first medium is described by the displacement equation $y = 0.02 \sin \left(100\pi t - \frac{4\pi}{3} x\right)$ , where $x$ and $y$ are in meters and $t$ is in seconds. As the wave enters a second medium, its speed becomes $120\text{ m/s}$ . What is the wavelength of the wave in the second medium?

2.4\text{ m}

1.5\text{ m}

0.42\text{ m}

3.0\text{ m}

Show answer & explanation

Answer: $2.4\text{ m}$

Answer

The wavelength of the wave in the second medium is

2.4\text{ m}

.

Comparing the wave equation

y = 0.02 \sin\left(100\pi t - \frac{4\pi}{3} x\right)

with

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

gives

\omega = 100\pi\text{ rad/s}

. The frequency is

f = \frac{\omega}{2\pi} = 50\text{ Hz}

. Because frequency does not change when crossing boundaries, the wavelength in the second medium where speed is

120\text{ m/s}

is

\lambda = \frac{v}{f} = \frac{120}{50} = 2.4\text{ m}

.

Step-by-Step Solution

1

Extract the angular frequency

\omega

from the given wave equation.

The angular frequency

\omega = 100\pi\text{ rad/s}

.

The standard wave equation is

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

.

2

Calculate the frequency of the wave.

f = \frac{\omega}{2\pi} = \frac{100\pi}{2\pi} = 50\text{ Hz}

.

Frequency is determined by the source and remains constant regardless of the medium.

3

Determine the wavelength in the second medium using the new wave speed.

\lambda_2 = \frac{v_2}{f} = \frac{120\text{ m/s}}{50\text{ Hz}} = 2.4\text{ m}

.

Applying the wave equation

v = f\lambda

with the updated speed in the second medium.

Key Concept

Invariance of wave frequency across media boundaries and extraction of wave parameters from the mathematical wave equation.

Question 83Question →

A sound wave of frequency $250\text{ Hz}$ propagates through a metal rod at a speed of $5000\text{ m s}^{-1}$ . What is the distance between a compression and the adjacent rarefaction in the rod?

20\text{ m}

0.05\text{ m}

10\text{ m}

1.25 \times 10^6\text{ m}

Show answer & explanation

Answer: $10\text{ m}$

Answer

The distance between a compression and the adjacent rarefaction is

10\text{ m}

.

First, find the wavelength using

\lambda = \frac{v}{f} = \frac{5000\text{ m s}^{-1}}{250\text{ Hz}} = 20\text{ m}

. In a longitudinal mechanical wave propagating through a medium, one complete wavelength is the distance between two successive compressions or two successive rarefactions. The distance from a compression to the adjacent rarefaction is half a wavelength, giving

\frac{20\text{ m}}{2} = 10\text{ m}

.

Step-by-Step Solution

1

Calculate the wavelength (\lambda) of the wave using the wave equation.

\lambda = \frac{v}{f} = \frac{5000\text{ m s}^{-1}}{250\text{ Hz}} = 20\text{ m}

The fundamental wave relationship is

v = f\lambda

, so wavelength is the ratio of wave speed to frequency.

2

Determine the distance between a compression and the adjacent rarefaction.

\text{Distance} = \frac{\lambda}{2} = \frac{20\text{ m}}{2} = 10\text{ m}

In a longitudinal wave, one full wavelength is the distance between two consecutive compressions. The distance between a compression and the immediate next rarefaction is half of one wavelength.

Key Concept

Distance between consecutive compression and rarefaction in a longitudinal wave

Question 84Question →

A mechanical longitudinal wave of frequency $250\text{ Hz}$ propagates from Medium 1 into Medium 2. In Medium 1, the distance between two consecutive compressions is $1.40\text{ m}$ . Upon entering Medium 2, the wave speed increases by $40\%$ . What is the distance, in meters, between a compression and the immediately adjacent rarefaction in Medium 2?

Show answer & explanation

Answer: 0.98

Answer

The distance between a compression and the adjacent rarefaction in Medium 2 is

0.98\text{ m}

.

When a mechanical wave travels between media, its frequency remains unchanged. The wave speed equation

v = f\lambda

indicates that wavelength is directly proportional to wave speed. A

40\%

increase in speed increases the wavelength in Medium 2 from

1.40\text{ m}

to

1.40 \times 1.40\text{ m} = 1.96\text{ m}

. In a longitudinal wave, the distance between a compression and an adjacent rarefaction is half a wavelength, yielding

\frac{1.96\text{ m}}{2} = 0.98\text{ m}

.

Step-by-Step Solution

1

Identify the wavelength in Medium 1 from the compression spacing.

\lambda_1 = 1.40\text{ m}

The distance between two successive compressions in a longitudinal wave corresponds to one complete wavelength.

2

Calculate the wavelength in Medium 2 using the constant frequency principle across media boundaries.

\lambda_2 = 1.40 \times 1.40\text{ m} = 1.96\text{ m}

The frequency of a wave is determined by the source and does not change upon entering a new medium. Since

v = f\lambda

, a

40\%

increase in wave speed results in a proportional

40\%

increase in wavelength.

3

Find the distance between a compression and the adjacent rarefaction in Medium 2.

d = \frac{\lambda_2}{2} = \frac{1.96\text{ m}}{2} = 0.98\text{ m}

In any longitudinal wave, a compression and its adjacent rarefaction are out of phase by half a cycle, corresponding to half a wavelength.

Key Concept

Wave propagation across boundaries and spatial separation of compressions and rarefactions in longitudinal waves.

Question 85Question →

A progressive wave traveling along a medium is described by the equation $y = 0.05 \sin(20\pi t - 4\pi x)$ , where $x$ and $y$ are measured in meters and $t$ is in seconds. What is the speed of the wave in $\text{m/s}$ ?

Show answer & explanation

Answer: 5

Answer

The speed of the wave is

5.0\text{ m/s}

.

Comparing

y = 0.05 \sin(20\pi t - 4\pi x)

to the standard wave equation

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

, we find

\omega = 20\pi\text{ rad/s}

and

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

. The wave speed

v

is calculated as

v = \frac{\omega}{k} = \frac{20\pi}{4\pi} = 5.0\text{ m/s}

.

Step-by-Step Solution

1

Identify the wave parameters from the standard equation form

\omega = 20\pi\text{ rad/s}

and

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

Matching the given equation

y = 0.05 \sin(20\pi t - 4\pi x)

to

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

gives the values for angular frequency

\omega

and wave number

k

.

2

Compute wave speed using the relationship

v = \frac{\omega}{k}

v = \frac{20\pi}{4\pi} = 5.0\text{ m/s}

Wave speed is defined as the ratio of angular frequency to wave number.

Key Concept

Wave Speed from Wave Equation

Question 86Question →

A plane mechanical wave propagating through a fluid is represented by the equation $y = 0.02 \sin \left(250\pi t - \frac{5\pi}{2} x\right)$ , where $x$ and $y$ are in meters and $t$ is in seconds. When the wave passes into a secondary fluid medium, its speed decreases to $60\text{ m/s}$ . Assuming the frequency remains constant, what is the wavelength of the wave in the secondary medium?

0.48\text{ m}

0.80\text{ m}

0.24\text{ m}

1.33\text{ m}

Show answer & explanation

Answer: $0.48\text{ m}$

Answer

The wavelength of the wave in the secondary medium is

0.48\text{ m}

.

Comparing the given equation to the general progressive wave equation

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

, we find the angular frequency

\omega = 250\pi\text{ rad/s}

. The wave frequency is

f = \frac{\omega}{2\pi} = 125\text{ Hz}

. Because wave frequency is invariant across media boundaries,

f

remains

125\text{ Hz}

in the secondary medium. Using the wave equation

v = f \lambda

, the wavelength in the secondary fluid is

\lambda = \frac{v}{f} = \frac{60\text{ m/s}}{125\text{ Hz}} = 0.48\text{ m}

.

Step-by-Step Solution

1

Extract angular frequency and wave number from the progressive wave equation

From

y = 0.02 \sin \left(250\pi t - \frac{5\pi}{2} x\right)

,

\omega = 250\pi\text{ rad/s}

and

k = \frac{5\pi}{2}\text{ rad/m}

.

Standard progressive wave equations follow the format

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

.

2

Determine the frequency of the wave

f = \frac{\omega}{2\pi} = \frac{250\pi}{2\pi} = 125\text{ Hz}

.

Frequency is related to angular frequency by

\omega = 2\pi f

.

3

Apply boundary transition rules to determine the new wavelength

Frequency

f

remains constant at

125\text{ Hz}

across media. Therefore,

\lambda_2 = \frac{v_2}{f} = \frac{60\text{ m/s}}{125\text{ Hz}} = 0.48\text{ m}

.

When a wave travels across different media boundaries, its frequency depends only on the source and remains constant, whereas speed and wavelength adjust accordingly.

Key Concept

Wave equation parameter extraction and frequency invariance during refraction

Estimated Time:1m 30s

Question 87Question →

Under which of the following conditions will total internal reflection occur when light encounters the boundary between two transparent media?

When light travels from a medium of higher refractive index to one of lower refractive index at an angle of incidence greater than the critical angle.

When light travels from a medium of lower refractive index to one of higher refractive index at an angle of incidence greater than the critical angle.

When light travels from a medium of higher refractive index to one of lower refractive index at an angle of incidence less than the critical angle.

When light travels from a medium of lower refractive index to one of higher refractive index at normal incidence.

Show answer & explanation

Answer: When light travels from a medium of higher refractive index to one of lower refractive index at an angle of incidence greater than the critical angle.

Answer

Total internal reflection occurs when light travels from a medium of higher refractive index to a medium of lower refractive index at an angle of incidence greater than the critical angle.

Total internal reflection takes place only when light moves from an optically denser medium (higher refractive index) into an optically less dense medium (lower refractive index), and the angle of incidence at the interface is greater than the critical angle for the two media.

Step-by-Step Solution

1

Identify the optical density requirement for total internal reflection

Light must travel from an optically denser medium (higher refractive index

n_1

) toward an optically rarer medium (lower refractive index

n_2

).

This condition ensures that the light refracts away from the normal into the second medium.

2

Identify the angular requirement at the interface

The angle of incidence

i

must be strictly greater than the critical angle

\theta_c

(

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

).

When

i > \theta_c

, no light can refract into the second medium, resulting in total reflection back into the initial medium.

Key Concept

Conditions for Total Internal Reflection

Question 88Question →

A progressive transverse wave traveling along a taut string is governed by the displacement equation $y(x,t) = 0.05 \sin(200\pi t - 10\pi x)$ , where $x$ and $y$ are measured in meters and $t$ in seconds. Calculate the distance, in meters, traveled by the wave front during the time taken for a single particle on the string to complete $15$ full oscillations.

Show answer & explanation

Answer: 3

Answer

The distance traveled by the wave front during 15 full particle oscillations is

3.0\text{ m}

.

Comparing

y(x,t) = 0.05 \sin(200\pi t - 10\pi x)

to the standard wave equation

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

gives

\omega = 200\pi\text{ rad/s}

and

k = 10\pi\text{ rad/m}

. The wave speed

v = \frac{\omega}{k} = 20\text{ m/s}

. The time for one full oscillation is

T = \frac{2\pi}{\omega} = 0.01\text{ s}

, so 15 full oscillations take

t = 15 \times 0.01\text{ s} = 0.15\text{ s}

. The distance traveled by the wave front is

d = v \times t = 20\text{ m/s} \times 0.15\text{ s} = 3.0\text{ m}

. Alternatively, because a wave travels a distance of one wavelength

\lambda = \frac{2\pi}{k} = 0.2\text{ m}

during each period (1 full oscillation), in 15 full oscillations the wave travels

15 \times \lambda = 15 \times 0.2\text{ m} = 3.0\text{ m}

.

Step-by-Step Solution

1

Extract wave parameters from the progressive wave equation

Angular frequency

\omega = 200\pi\text{ rad/s}

and wave number

k = 10\pi\text{ rad/m}

.

Matching the given equation

y = 0.05 \sin(200\pi t - 10\pi x)

with the standard form

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

identifies

\omega

and

k

.

2

Calculate the wave propagation velocity

v = 20\text{ m/s}

.

Wave speed is given by the relation

v = \frac{\omega}{k} = \frac{200\pi}{10\pi} = 20\text{ m/s}

.

3

Find the period of oscillation and total elapsed time

Period

T = 0.01\text{ s}

, total time

t = 0.15\text{ s}

.

The period

T = \frac{2\pi}{\omega} = \frac{2\pi}{200\pi} = 0.01\text{ s}

. For 15 complete oscillations,

t = 15 \times 0.01\text{ s} = 0.15\text{ s}

.

4

Compute the total distance traveled by the wave

Distance

d = 3.0\text{ m}

.

Using linear motion at constant wave speed,

d = v \times t = 20\text{ m/s} \times 0.15\text{ s} = 3.0\text{ m}

.

Key Concept

Wave equation parameters, particle oscillation period, and wave propagation distance

Estimated Time:2m 0s

Question 89Question →

Match each type of wave listed on the left with its correct classification and propagation characteristic on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Sound wave in air matches Mechanical longitudinal wave requiring a material medium; Radio wave in vacuum matches Electromagnetic transverse wave capable of traveling without a medium; Water ripple on a lake surface matches Mechanical transverse wave propagating along a liquid surface.

Each wave is correctly paired based on whether it needs a physical medium to propagate (mechanical waves require a medium, electromagnetic waves do not) and whether the displacement is parallel (longitudinal) or perpendicular (transverse) to the direction of energy propagation.

Step-by-Step Solution

1

Identify the medium requirement and vibration direction for a sound wave in air.

Sound waves require a material medium (air) and vibrate parallel to the direction of wave movement, making them mechanical longitudinal waves.

Classification depends on whether a physical medium is needed and how particles oscillate relative to energy transport.

2

Identify the medium requirement and vibration direction for a radio wave in a vacuum.

Radio waves can travel through empty space without a material medium and consist of field oscillations perpendicular to propagation, making them electromagnetic transverse waves.

Electromagnetic waves propagate via mutually perpendicular electric and magnetic field oscillations and require no medium.

3

Identify the medium requirement and vibration direction for a water ripple.

Ripples require a material medium (water) and displace surface water up and down perpendicular to wave travel, making them mechanical transverse waves.

Surface water waves exhibit transverse displacement characteristics in a physical liquid medium.

Key Concept

Classification of waves based on medium requirement (mechanical vs. electromagnetic) and particle vibration direction relative to propagation (transverse vs. longitudinal).

Question 90Question →

A harmonic wave traveling through an initial elastic medium is represented by the wave equation $y = 0.04 \cos\left(50\pi t - \frac{\pi}{4} x\right)$ , where $x$ and $y$ are measured in meters and $t$ is in seconds. When this wave passes into a second medium, its propagation speed doubles. What is the wavelength of the wave in the second medium?

4.0\text{ m}

8.0\text{ m}

16.0\text{ m}

32.0\text{ m}

Show answer & explanation

Answer: $16.0\text{ m}$

Answer

The wavelength of the wave in the second medium is

16.0\text{ m}

.

The wave's angular frequency

\omega = 50\pi\text{ rad/s}

corresponds to a source frequency of

f = 25\text{ Hz}

, and its wave number

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

corresponds to an initial wavelength

\lambda_1 = 8.0\text{ m}

. The wave speed in the first medium is

v_1 = f \lambda_1 = 200\text{ m/s}

. In the second medium, the wave speed doubles to

v_2 = 400\text{ m/s}

. Because frequency is determined by the source and remains invariant during refraction across media boundaries (

f_2 = f_1 = 25\text{ Hz}

), the new wavelength becomes

\lambda_2 = \frac{v_2}{f} = \frac{400\text{ m/s}}{25\text{ Hz}} = 16.0\text{ m}

.

Step-by-Step Solution

1

Extract angular frequency

\omega

and wave number

k

from the general wave equation

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

.

\omega = 50\pi\text{ rad/s}

and

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

.

Matching coefficients in the standard wave equation provides the temporal and spatial frequencies of the wave.

2

Calculate the frequency

f

and wavelength

\lambda_1

in the first medium.

f = \frac{\omega}{2\pi} = \frac{50\pi}{2\pi} = 25\text{ Hz}

and

\lambda_1 = \frac{2\pi}{k} = \frac{2\pi}{\pi/4} = 8.0\text{ m}

.

Fundamental wave relationships connect angular frequency to frequency and wave number to wavelength.

3

Determine the wave speed

v_1

in the first medium and

v_2

in the second medium.

v_1 = f \lambda_1 = 25 \times 8.0 = 200\text{ m/s}

. Therefore,

v_2 = 2 \times v_1 = 400\text{ m/s}

.

The problem states that wave propagation speed doubles upon entering the second medium.

4

Calculate the wavelength

\lambda_2

in the second medium using constant frequency

f = 25\text{ Hz}

.

\lambda_2 = \frac{v_2}{f} = \frac{400}{25} = 16.0\text{ m}

.

When a wave crosses a boundary between two media, its frequency is determined solely by the source and remains constant.

Key Concept

Frequency invariance across media boundaries and the wave speed equation

v = f \lambda

Estimated Time:1m 30s

Question 91Question →

A ray of light traveling inside a transparent medium strikes the boundary with air. If the critical angle for total internal reflection at this boundary is $30^\circ$ , what is the refractive index of the medium?

2.00

0.50

0.87

1.15

Show answer & explanation

Answer: $2.00$

Answer

The refractive index of the medium is

2.00

.

For light traveling from a medium into air, the critical angle

C

is related to the refractive index

n

by

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

. Substituting

C = 30^\circ

gives

n = \frac{1}{\sin 30^\circ} = \frac{1}{0.5} = 2.00

.

Step-by-Step Solution

1

Identify the relationship between critical angle

C

and refractive index

n

The formula for light passing into air is

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

.

Total internal reflection occurs when light travels from a denser medium to a less dense medium (air,

n_{air} = 1

) at an angle greater than the critical angle.

2

Substitute the given critical angle

C = 30^\circ

into the formula

\sin 30^\circ = 0.5

, so

n = \frac{1}{0.5} = 2.00

.

Taking the reciprocal of

\sin 30^\circ

gives the correct refractive index.

Key Concept

Critical Angle and Refractive Index

Question 92Question →

A water wave with a frequency of $20\text{ Hz}$ and a wavelength of $0.60\text{ m}$ in deep water enters a shallow region where its speed becomes $8.0\text{ m s}^{-1}$ . What is the wavelength of the wave in the shallow region?

0.40\text{ m}

0.60\text{ m}

0.90\text{ m}

1.50\text{ m}

Show answer & explanation

Answer: $0.40\text{ m}$

Answer

The wavelength of the wave in the shallow region is

0.40\text{ m}

.

When a wave passes from one medium into another, its frequency remains constant because frequency depends only on the source. Applying the wave equation

v = f\lambda

to the shallow region gives

\lambda = \frac{v}{f} = \frac{8.0\text{ m s}^{-1}}{20\text{ Hz}} = 0.40\text{ m}

.

Step-by-Step Solution

1

Identify the constant parameter during wave refraction.

The frequency of the wave remains

f = 20\text{ Hz}

.

Frequency is determined entirely by the source producing the wave and does not change upon entering a new medium.

2

Calculate the new wavelength using the wave equation

v = f\lambda

.

$\lambda = \frac{v}{f} = \frac{8.0\text{ m s}^{-1}}{20\text{ Hz}} = 0.40\text{ m}$.

Dividing the wave speed in the new medium by the constant frequency yields the wavelength in that medium.

Key Concept

Constancy of wave frequency during refraction across media boundaries

Question 93Question →

In a double-slit experiment, monochromatic light of wavelength $500\text{ nm}$ is incident normally on two narrow slits. If the third-order ( $m = 3$ ) bright fringe is observed at an angle of $30^\circ$ from the central maximum, what is the slit separation, $d$ , in micrometers ( $\mu\text{m}$ )?

Show answer & explanation

Answer: 3

Answer

The slit separation is

3.0\text{ }\mu\text{m}

.

For bright fringes in a double-slit setup, constructive interference occurs when

d \sin\theta = m\lambda

. Rearranging for slit separation yields

d = \frac{m\lambda}{\sin\theta}

. Substituting

m = 3

,

\lambda = 0.50\text{ }\mu\text{m}

, and

\theta = 30^\circ

gives

d = \frac{3 \times 0.50\text{ }\mu\text{m}}{\sin 30^\circ} = \frac{1.50\text{ }\mu\text{m}}{0.50} = 3.0\text{ }\mu\text{m}

.

Step-by-Step Solution

1

Identify the constructive interference equation for Young's double-slit experiment.

d \sin\theta = m\lambda

Bright fringes occur where waves from the two slits interfere constructively, which corresponds to path differences equal to integer multiples of the wavelength.

2

Convert the given wavelength to micrometers and substitute known values.

\lambda = 0.50\text{ }\mu\text{m}

,

m = 3

,

\sin(30^\circ) = 0.50

Converting units to micrometers early simplifies direct calculation of

d

in

\mu\text{m}

.

3

Rearrange the equation and evaluate for

d

.

d = \frac{3 \times 0.50\text{ }\mu\text{m}}{0.50} = 3.0\text{ }\mu\text{m}

Dividing the numerator by

\sin(30^\circ) = 0.50

doubles the value of

m\lambda

.

Key Concept

Angular position condition for constructive interference in double-slit diffraction

Question 94Question →

A progressive wave traveling through a uniform medium is described by the displacement equation $y = 0.04 \sin\left(150\pi t - 6\pi x\right)$ , where $x$ and $y$ are measured in meters and $t$ is in seconds. What is the speed of propagation of the wave?

Show answer & explanation

Answer: 25

Answer

The speed of propagation of the wave is 25 m/s.

By matching

y = 0.04 \sin(150\pi t - 6\pi x)

to the standard wave equation

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

, we obtain

\omega = 150\pi\text{ rad/s}

and

k = 6\pi\text{ rad/m}

. The speed of the wave

v

is calculated as

v = \frac{\omega}{k} = \frac{150\pi}{6\pi} = 25\text{ m/s}

.

Step-by-Step Solution

1

Compare the given wave equation with the general progressive wave equation

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

.

Angular frequency

\omega = 150\pi\text{ rad/s}

and wave number

k = 6\pi\text{ rad/m}

.

Matching coefficients allows direct extraction of angular frequency and spatial wave number.

2

Calculate the wave speed using the relationship

v = \frac{\omega}{k}

.

v = \frac{150\pi}{6\pi} = 25\text{ m/s}

.

The velocity of a progressive wave is equal to the ratio of its angular frequency to its wave number.

Key Concept

Determining wave velocity from progressive wave equation parameters

Question 95Question →

A mechanical wave propagating along a string is defined by the displacement equation $y = 0.05 \sin\left(160\pi t - 8\pi x + \frac{\pi}{3}\right)$ , where $x$ and $y$ are in meters and $t$ is in seconds. The wave transitions into a different section of string where its propagation speed drops by $25\%$ . Calculate the minimum distance (in meters) between two points in this second section that have a phase difference of $\frac{2\pi}{3}\text{ rad}$ .

Show answer & explanation

Answer: 0.0625

Answer

The minimum distance between the two points in the second section is

0.0625\text{ m}

.

The correct calculation gives

0.0625\text{ m}

. Comparing

y = 0.05 \sin(160\pi t - 8\pi x + \pi/3)

with the general form

y = A \sin(\omega t - k x + \phi_0)

identifies

\omega = 160\pi\text{ rad/s}

and

k_1 = 8\pi\text{ rad/m}

, yielding an initial speed of

v_1 = \omega / k_1 = 20\text{ m/s}

. Upon transitioning into the second string section, the speed drops by

25\%

to

v_2 = 15\text{ m/s}

. Since the angular frequency

\omega = 160\pi\text{ rad/s}

remains invariant during refraction, the wave number in the second section is

k_2 = \omega / v_2 = 160\pi / 15 = 32\pi / 3\text{ rad/m}

. Substituting

k_2

and the given phase difference

\Delta \phi = 2\pi / 3\text{ rad}

into

\Delta \phi = k_2 \Delta x

yields

\Delta x = (2\pi / 3) / (32\pi / 3) = 2/32 = 0.0625\text{ m}

.

Step-by-Step Solution

1

Extract angular frequency and wave number from the displacement equation

\omega = 160\pi\text{ rad/s}

and

k_1 = 8\pi\text{ rad/m}

The standard progressive wave equation is formatted as

y = A \sin(\omega t - k x + \phi_0)

.

2

Calculate the initial wave propagation speed

v_1

v_1 = \frac{\omega}{k_1} = \frac{160\pi}{8\pi} = 20\text{ m/s}

Wave speed is equal to the ratio of angular frequency to wave number.

3

Determine the wave speed

v_2

in the second section

v_2 = 20 \times (1 - 0.25) = 15\text{ m/s}

The wave speed decreases by

25\%

, making

v_2 = 0.75 v_1

.

4

Find the new wave number

k_2

in the second section

k_2 = \frac{\omega}{v_2} = \frac{160\pi}{15} = \frac{32\pi}{3}\text{ rad/m}

Frequency and angular frequency remain invariant when a wave passes from one medium to another.

5

Calculate the spatial separation

\Delta x

for the specified phase difference

\Delta x = \frac{\Delta \phi}{k_2} = \frac{2\pi / 3}{32\pi / 3} = \frac{2}{32} = 0.0625\text{ m}

Phase difference relates to spatial distance via

\Delta \phi = k \Delta x

.

Key Concept

Wave Equation Parameter Extraction and Invariance of Frequency in Refraction

Question 96Question →

A progressive transverse wave traveling along a medium is described by the displacement equation $y = 0.05 \sin\left(160\pi t - 4\pi x\right)$ , where $x$ and $y$ are in meters and $t$ is in seconds. What is the ratio of the maximum particle velocity to the wave propagation velocity? (Take $\pi = 3.142$ .)

Show answer & explanation

Answer: 0.628

Answer

The ratio of the maximum particle velocity to the wave propagation velocity is 0.628.

For a progressive wave

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

, the wave advances at speed

v = \frac{\omega}{k} = \frac{160\pi}{4\pi} = 40\text{ m/s}

. Meanwhile, individual particles vibrate transversely with simple harmonic motion where maximum velocity is

v_{p,\text{max}} = A\omega = 0.05 \times 160\pi = 8\pi\text{ m/s}

. Taking the ratio gives

\frac{v_{p,\text{max}}}{v} = \frac{8\pi}{40} = 0.2\pi = 0.2 \times 3.142 = 0.6284

, which rounds to 0.628.

Step-by-Step Solution

1

Extract parameters from the standard wave equation

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

A = 0.05\text{ m}

,

\omega = 160\pi\text{ rad/s}

,

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

Matching the given equation with standard progressive wave form yields the required wave constants.

2

Calculate the wave propagation velocity

v

v = \frac{\omega}{k} = \frac{160\pi}{4\pi} = 40\text{ m/s}

The speed at which the wave energy advances through the medium depends on angular frequency and wave number.

3

Calculate the maximum transverse particle velocity

v_{p,\text{max}}

v_{p,\text{max}} = A\omega = 0.05 \times 160\pi = 8\pi\text{ m/s} \approx 25.136\text{ m/s}

Particles perform simple harmonic motion, whose maximum speed is given by the product of amplitude and angular frequency.

4

Compute the ratio of maximum particle velocity to wave velocity

Ratio

= \frac{v_{p,\text{max}}}{v} = \frac{8\pi}{40} = 0.2\pi = 0.2 \times 3.142 = 0.6284

Dividing the maximum particle speed by the wave speed gives the desired dimensionless ratio.

Key Concept

Distinction between particle oscillation velocity and wave propagation velocity

Question 97Question →

A transverse wave of frequency $15\text{ Hz}$ propagates from a dense medium where its speed is $45\text{ m s}^{-1}$ into a lighter medium where its speed decreases to $30\text{ m s}^{-1}$ . What is the frequency of the wave in the second medium?

10\text{ Hz}

15\text{ Hz}

22.5\text{ Hz}

2.0\text{ Hz}

Show answer & explanation

Answer: $15\text{ Hz}$

Answer

The frequency of the wave in the second medium is

15\text{ Hz}

.

The correct answer is

15\text{ Hz}

because the frequency of a wave is determined solely by the source that creates it. When a wave passes from one medium into another, its speed and wavelength change proportionally, but the frequency remains constant.

Step-by-Step Solution

1

Identify the property of a wave that is source-dependent and invariant across boundaries.

The frequency (

f

) of a wave is determined entirely by the oscillating source generating the wave.

When a wave crosses the boundary between two different media, its speed (

v

) and wavelength (

λ

) change due to the properties of the new medium, but its frequency (

f

) remains constant.

2

Determine the frequency in the new medium.

f_{\text{second medium}} = f_{\text{initial}} = 15\text{ Hz}

.

Since frequency does not change across medium boundaries, the wave maintains a frequency of

15\text{ Hz}

.

Key Concept

Invariance of Wave Frequency Across Media Boundaries

Estimated Time:45s

Question 98Question →

A progressive wave propagating through an initial medium is represented by the displacement equation $y = 0.05 \sin\left(100\pi t - \frac{2\pi}{5} x\right)$ , where $x$ and $y$ are in metres and $t$ is in seconds. Upon entering a second medium, the wave speed decreases by $20\%$ . What is the wavelength of the wave in the second medium?

4.0\text{ m}

5.0\text{ m}

6.25\text{ m}

2.0\text{ m}

Show answer & explanation

Answer: $4.0\text{ m}$

Answer

4.0\text{ m}

Comparing

y = 0.05 \sin\left(100\pi t - \frac{2\pi}{5} x\right)

with the standard wave equation

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

yields a wave number

k = \frac{2\pi}{5}\text{ rad/m}

, giving an initial wavelength

\lambda_1 = \frac{2\pi}{k} = 5.0\text{ m}

. When a wave refracts into another medium, its frequency stays constant, making wave speed and wavelength directly proportional. A

20\%

reduction in speed decreases the wavelength by

20\%

, giving

\lambda_2 = 5.0 \times 0.80 = 4.0\text{ m}

.

Step-by-Step Solution

1

Extract the angular frequency

\omega

and wave number

k

from the wave equation

\omega = 100\pi\text{ rad/s}

and

k = \frac{2\pi}{5}\text{ rad/m}

The standard progressive wave equation is

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

2

Calculate the wavelength

\lambda_1

in the first medium

$\lambda_1 = \frac{2\pi}{k} = \frac{2\pi}{\frac{2\pi}{5}} = 5.0\text{ m}$

The wave number is related to wavelength by

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

3

Determine the new wavelength

\lambda_2

in the second medium using the refraction property

$\lambda_2 = \lambda_1 \times (1 - 0.20) = 5.0 \times 0.80 = 4.0\text{ m}$

When a wave crosses a boundary between two media, its frequency

f

remains constant. Therefore, wave speed

v = f\lambda

is directly proportional to wavelength

\lambda

Key Concept

Constancy of wave frequency during refraction and wave equation parameters extraction

Estimated Time:2m 0s

Question 99Question →

A periodic water wave of frequency $5.0\text{ Hz}$ has a wavelength of $1.2\text{ m}$ in deep water. When the wave enters a shallow section of the ripple tank, its speed reduces to $4.0\text{ m s}^{-1}$ . What is the wavelength of the wave in the shallow section?

0.67\text{ m}

0.80\text{ m}

1.5\text{ m}

1.8\text{ m}

Show answer & explanation

Answer: $0.80\text{ m}$

Answer

0.80\text{ m}

When a wave travels from one medium to another (e.g., deep to shallow water), its frequency

f

remains constant because frequency is dependent only on the wave source. Using the wave equation

v = f\lambda

, the wavelength in the shallow water is calculated as

\lambda = \frac{v}{f} = \frac{4.0\text{ m s}^{-1}}{5.0\text{ Hz}} = 0.80\text{ m}

.

Step-by-Step Solution

1

Determine the invariant property across media boundaries

The frequency

f

remains constant at

5.0\text{ Hz}

when a wave passes from deep water to shallow water.

Frequency is determined solely by the source of the wave vibration, not the medium of propagation.

2

Calculate the wavelength in the shallow section using the wave equation

\lambda_2 = \frac{v_2}{f} = \frac{4.0\text{ m s}^{-1}}{5.0\text{ Hz}} = 0.80\text{ m}

Applying the relationship

v = f\lambda

for the second medium.

Key Concept

Frequency invariance of waves across boundaries and application of the wave equation

v = f\lambda

Question 100Question →

A girl stands at a specific distance from a flat vertical wall and claps her hands once. If she hears the echo $0.4\text{ s}$ later, what is her distance from the wall in meters? (Take the speed of sound in air as $340\text{ m/s}$ )

Show answer & explanation

Answer: 68

Answer

The distance of the girl from the wall is

68\text{ m}

.

An echo involves the sound traveling to the reflecting surface and returning to the source, covering a total distance of

2d

. Using

2d = v \times t

, we obtain

d = \frac{340 \times 0.4}{2} = 68\text{ m}

.

Step-by-Step Solution

1

Identify the given values from the problem statement.

Time for echo

t = 0.4\text{ s}

, speed of sound

v = 340\text{ m/s}

.

An echo is a reflected sound wave that travels to the wall and back, completing a round trip.

2

Apply the echo calculation formula.

d = \frac{v \times t}{2}

The total distance traveled by the sound is

2d

. Thus, the one-way distance

d

to the reflecting surface is half of the total distance.

3

Substitute the values and calculate the distance.

d = \frac{340 \times 0.4}{2} = 68\text{ m}

Multiplying the speed by half the elapsed time gives the distance to the wall.

Key Concept

Echo distance calculation

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