Simple Harmonic Motion Practice Questions — JAMB UTME

Question 1Question →

A body of mass $0.4\text{ kg}$ suspended vertically from a helical spring produces a static extension of $0.1\text{ m}$ . The body is then pulled down further and set into vertical simple harmonic motion with an amplitude of $0.05\text{ m}$ . What is the maximum velocity of the body in $\text{m/s}$ ? (Take $g = 10\text{ m/s}^2$ )

Show answer & explanation

Answer: 0.5

Answer

The maximum velocity of the body during oscillation is

0.5\text{ m/s}

.

At static equilibrium, weight balances restoring force (

mg = ke

), giving

\frac{k}{m} = \frac{g}{e} = \frac{10}{0.1} = 100\text{ s}^{-2}

. The angular frequency is

\omega = \sqrt{\frac{k}{m}} = 10\text{ rad/s}

. In SHM, the maximum velocity occurs at the central equilibrium position and is given by

v_{\max} = \omega A = 10 \times 0.05 = 0.5\text{ m/s}

.

Step-by-Step Solution

1

Relate spring stiffness to static extension

\frac{k}{m} = 100\text{ s}^{-2}

At vertical static equilibrium, the weight of the mass equals the restoring force:

mg = ke \implies \frac{k}{m} = \frac{g}{e} = \frac{10\text{ m/s}^2}{0.1\text{ m}} = 100\text{ s}^{-2}

.

2

Determine the angular frequency

\omega = 10\text{ rad/s}

The angular frequency of a mass-spring system is given by

\omega = \sqrt{\frac{k}{m}} = \sqrt{100} = 10\text{ rad/s}

.

3

Calculate maximum velocity

v_{\max} = 0.5\text{ m/s}

The maximum speed in simple harmonic motion occurs at the equilibrium position and is computed using

v_{\max} = \omega A = 10\text{ rad/s} \times 0.05\text{ m} = 0.5\text{ m/s}

.

Key Concept

Maximum velocity and angular frequency derived from static extension in Simple Harmonic Motion

Question 2Question →

A simple pendulum of length $l$ has a period of oscillation of $2.0\text{ s}$ when suspended with a bob of mass $50\text{ g}$ . If the bob is replaced with one of mass $200\text{ g}$ while maintaining the exact same string length, what is the new period of oscillation?

2.0\text{ s}

4.0\text{ s}

1.0\text{ s}

8.0\text{ s}

Show answer & explanation

Answer: $2.0\text{ s}$

Answer

The period of oscillation remains

2.0\text{ s}

.

The period of oscillation of a simple pendulum undergoing simple harmonic motion is given by

T = 2\pi \sqrt{\frac{l}{g}}

, where

l

is the length of the pendulum and

g

is the acceleration due to gravity. Because the mass of the bob does not appear in this equation, changing the mass from

50\text{ g}

to

200\text{ g}

does not alter the period. Therefore, the period remains

2.0\text{ s}

.

Step-by-Step Solution

1

Identify the formula for the period of a simple pendulum.

T = 2\pi \sqrt{\frac{l}{g}}

The period

T

depends on the length of the pendulum

l

and local acceleration due to gravity

g

.

2

Analyze the dependence of period on bob mass.

The mass parameter

m

does not appear in the formula

T = 2\pi \sqrt{\frac{l}{g}}

.

The period of a simple pendulum is independent of the mass of the bob.

3

Determine the new period of oscillation.

T_{new} = 2.0\text{ s}

Since length

l

and acceleration due to gravity

g

remain unchanged, the period stays

2.0\text{ s}

.

Key Concept

Independence of simple pendulum period from bob mass

Question 3Question →

A body of mass $0.2\text{ kg}$ undergoes simple harmonic motion with an angular frequency of $10\text{ rad/s}$ and an amplitude of $0.04\text{ m}$ . What is the magnitude of the maximum restoring force acting on the body in newtons?

Show answer & explanation

Answer: 0.8

Answer

The magnitude of the maximum restoring force acting on the body is

0.8\text{ N}

.

The maximum restoring force in simple harmonic motion occurs at maximum displacement (

y = A

) and is given by

F_{\text{max}} = m \omega^2 A

. Substituting

m = 0.2\text{ kg}

,

\omega = 10\text{ rad/s}

, and

A = 0.04\text{ m}

yields

F_{\text{max}} = 0.2 \times (10)^2 \times 0.04 = 0.8\text{ N}

.

Step-by-Step Solution

1

Identify the given physical quantities from the problem statement.

m = 0.2\text{ kg}

,

\omega = 10\text{ rad/s}

, and

A = 0.04\text{ m}

.

These parameters are required to calculate acceleration and restoring force in simple harmonic motion.

2

Calculate the maximum acceleration of the oscillating body.

a_{\text{max}} = \omega^2 A = (10)^2 \times 0.04 = 4.0\text{ m/s}^2

.

In simple harmonic motion, maximum acceleration occurs at maximum displacement (the amplitude).

3

Determine the maximum restoring force.

F_{\text{max}} = m a_{\text{max}} = 0.2 \times 4.0 = 0.8\text{ N}

.

According to Newton's second law, force is the product of mass and acceleration.

Key Concept

Maximum Restoring Force in Simple Harmonic Motion

Question 4Question →

A simple pendulum suspended in a terrestrial laboratory has a period of oscillation $T$ when fitted with a bob of mass $m$ . If the bob is replaced by another bob of mass $4m$ and the entire setup is moved to a high-altitude station where the acceleration due to gravity is $\frac{g}{4}$ , what is the new period of oscillation of the pendulum?

2T

4T

8T

\frac{T}{2}

Show answer & explanation

Answer: $2T$

Answer

The new period of oscillation is

2T

.

The period of a simple pendulum is given by

T = 2\pi \sqrt{\frac{L}{g}}

. It depends strictly on the length of the string

L

and the local gravitational acceleration

g

, making it independent of the bob's mass

m

. Replacing mass

m

with

4m

does not alter the period. When gravity decreases to

g' = \frac{g}{4}

, the new period becomes

T' = 2\pi \sqrt{\frac{L}{g/4}} = 2 \left(2\pi \sqrt{\frac{L}{g}}\right) = 2T

.

Step-by-Step Solution

1

Identify the formula for the period of a simple pendulum

The period

T

of a simple pendulum is given by

T = 2\pi \sqrt{\frac{L}{g}}

, where

L

is the pendulum length and

g

is the acceleration due to gravity.

This formula defines how physical parameters affect the oscillatory period of a pendulum under simple harmonic motion.

2

Analyze the effect of changing the bob's mass

Changing the mass of the bob from

m

to

4m

has no effect on the period.

The equation for the period of a simple pendulum contains no mass term, demonstrating that mass does not influence the period of small-angle oscillations.

3

Calculate the new period

T'

under the altered gravitational field

g' = \frac{g}{4}

T' = 2\pi \sqrt{\frac{L}{g/4}} = 2\pi \sqrt{\frac{4L}{g}} = 2 \times \left(2\pi \sqrt{\frac{L}{g}}\right) = 2T

.

Reducing the gravitational acceleration to one-fourth increases the square-root term

\sqrt{\frac{1}{1/4}} = 2

, thereby doubling the period.

Key Concept

Mass Independence and Gravity Dependence of Simple Pendulum Period

Question 5Question →

A particle executes simple harmonic motion along a straight line with an amplitude of $0.10\text{ m}$ . At what displacement from the equilibrium position is the kinetic energy of the particle equal to three times its potential energy?

0.05\text{ m}

0.087\text{ m}

0.071\text{ m}

0.025\text{ m}

Show answer & explanation

Answer: $0.05\text{ m}$

Answer

The displacement from the equilibrium position is

0.05\text{ m}

.

In simple harmonic motion, potential energy is

U = \frac{1}{2}kx^2

and kinetic energy is

K = \frac{1}{2}k(A^2 - x^2)

. Equating

K = 3U

yields

A^2 - x^2 = 3x^2

, which simplifies to

4x^2 = A^2

or

x = \frac{A}{2}

. For an amplitude of

0.10\text{ m}

, the displacement is

0.05\text{ m}

.

Step-by-Step Solution

1

Write the expressions for kinetic energy

K

and potential energy

U

in simple harmonic motion.

U = \frac{1}{2} k x^2

and

K = \frac{1}{2} k (A^2 - x^2)

, where

A

is amplitude and

x

is displacement.

These equations express the energy distribution at any displacement

x

.

2

Set up the condition given in the problem,

K = 3U

.

\frac{1}{2} k (A^2 - x^2) = 3 \times \left(\frac{1}{2} k x^2\right) \implies A^2 - x^2 = 3x^2

.

Canceling common factor

\frac{1}{2} k

simplifies the relationship between amplitude and displacement.

3

Solve for displacement

x

in terms of amplitude

A

.

A^2 = 4x^2 \implies x = \frac{A}{2}

.

Taking the square root of both sides gives the position where kinetic energy is three times potential energy.

4

Substitute the given amplitude

A = 0.10\text{ m}

to find

x

.

x = \frac{0.10\text{ m}}{2} = 0.05\text{ m}

.

Carrying out the calculation yields the final numerical displacement.

Key Concept

Conservation of Energy in Simple Harmonic Motion

Estimated Time:1m 15s

Question 6Question →

A $0.50\text{ kg}$ mass attached to a horizontal spring undergoes simple harmonic motion on a frictionless surface. The total mechanical energy of the system is $0.16\text{ J}$ and the force constant of the spring is $32\text{ N/m}$ . What is the speed of the mass, in $\text{m/s}$ , at the instant when the magnitude of its acceleration is $3.84\text{ m/s}^2$ ?

Show answer & explanation

Answer: 0.64

Answer

The speed of the mass at that instant is

0.64\text{ m/s}

.

Using the relation for total mechanical energy

E = \frac{1}{2}kA^2

, the amplitude is

A = 0.10\text{ m}

. The angular frequency is

\omega = \sqrt{k/m} = 8.0\text{ rad/s}

. From

|a| = \omega^2 |x|

, the displacement magnitude when acceleration is

3.84\text{ m/s}^2

is

|x| = 0.06\text{ m}

. Substituting these values into

v = \omega \sqrt{A^2 - x^2}

yields

v = 8.0 \sqrt{0.10^2 - 0.06^2} = 0.64\text{ m/s}

.

Step-by-Step Solution

1

Calculate the angular frequency of the simple harmonic motion

\omega = 8.0\text{ rad/s}

The angular frequency depends on the stiffness constant and the mass according to \omega = \sqrt{k/m}.

2

Calculate the amplitude of oscillation from total energy

A = 0.10\text{ m}

The total mechanical energy in SHM is given by E = \frac{1}{2}kA^2.

3

Find the magnitude of displacement corresponding to the given acceleration

|x| = 0.06\text{ m}

In SHM, acceleration magnitude is related to displacement magnitude by |a| = \omega^2 |x|.

4

Calculate the speed at this displacement using the SHM velocity-displacement relation

v = 0.64\text{ m/s}

Velocity in SHM is calculated using v = \omega \sqrt{A^2 - x^2}.

Key Concept

Interdependence of energy, angular frequency, acceleration, and velocity in Simple Harmonic Motion

Question 7Question →

An object of mass $0.20\text{ kg}$ executes simple harmonic motion with an amplitude of $0.05\text{ m}$ and a maximum acceleration of $20\text{ m/s}^2$ . What is the speed of the object when its displacement from the equilibrium position is $0.03\text{ m}$ ?

0.16\text{ m/s}

0.60\text{ m/s}

0.80\text{ m/s}

1.00\text{ m/s}

Show answer & explanation

Answer: $0.80\text{ m/s}$

Answer

The speed of the object at a displacement of

0.03\text{ m}

is

0.80\text{ m/s}

.

The maximum acceleration in simple harmonic motion is given by

a_{\text{max}} = \omega^2 A

. Substituting

a_{\text{max}} = 20\text{ m/s}^2

and

A = 0.05\text{ m}

gives

\omega^2 = 400\text{ rad}^2/\text{s}^2

, so

\omega = 20\text{ rad/s}

. The speed at any displacement

x

is given by

v = \omega \sqrt{A^2 - x^2}

. For

x = 0.03\text{ m}

,

v = 20 \sqrt{0.05^2 - 0.03^2} = 20 \times 0.04 = 0.80\text{ m/s}

.

Step-by-Step Solution

1

Determine the angular frequency (

\omega

) of the simple harmonic motion from the maximum acceleration formula.

\omega = 20\text{ rad/s}

Maximum acceleration is given by

a_{\text{max}} = \omega^2 A

. Rearranging gives

\omega^2 = \frac{a_{\text{max}}}{A} = \frac{20}{0.05} = 400\text{ rad}^2/\text{s}^2

, so

\omega = 20\text{ rad/s}

.

2

Calculate the speed (

v

) at the given displacement (

x = 0.03\text{ m}

) using the SHM velocity formula.

v = 0.80\text{ m/s}

The speed at displacement

x

is

v = \omega \sqrt{A^2 - x^2} = 20 \times \sqrt{0.05^2 - 0.03^2} = 20 \times \sqrt{0.0016} = 20 \times 0.04 = 0.80\text{ m/s}

.

Key Concept

Simple Harmonic Motion Velocity and Acceleration Relationships

Estimated Time:1m 30s

Question 8Question →

The maximum acceleration of a body oscillating in simple harmonic motion is $8\text{ m/s}^2$ . If the period of oscillation is $\pi\text{ s}$ , calculate the amplitude of the oscillation in meters.

Show answer & explanation

Answer: 2

Answer

The amplitude of the oscillation is

2.0\text{ m}

.

The correct answer of

2.0\text{ m}

is obtained by first deriving the angular frequency

\omega = \frac{2\pi}{T} = 2\text{ rad/s}

, and then using the relation

a_{\text{max}} = \omega^2 A

to solve for amplitude:

A = \frac{8}{2^2} = 2.0\text{ m}

.

Step-by-Step Solution

1

Calculate angular frequency (

\omega

) from the given period (

T

).

\omega = \frac{2\pi}{T} = \frac{2\pi}{\pi} = 2\text{ rad/s}

Angular frequency specifies the rate of phase change in oscillations.

2

Apply the maximum acceleration formula for simple harmonic motion to determine amplitude (

A

).

a_{\text{max}} = \omega^2 A \implies 8 = 2^2 \times A \implies A = 2.0\text{ m}

In simple harmonic motion, maximum acceleration occurs at the extreme position and equals

\omega^2 A

.

Key Concept

Simple Harmonic Motion Acceleration and Period Relationship

Question 9Question →

A simple pendulum of length $L$ with a bob of mass $m$ has a period of oscillation of $2.0\text{ s}$ . If the mass of the bob is increased to $4m$ and the length of the pendulum string is increased to $4L$ , what is the new period of oscillation?

2.0\text{ s}

4.0\text{ s}

8.0\text{ s}

16.0\text{ s}

Show answer & explanation

Answer: $4.0\text{ s}$

Answer

The new period of oscillation is

4.0\text{ s}

.

The period of a simple pendulum undergoing simple harmonic motion is given by

T = 2\pi \sqrt{\frac{L}{g}}

. The mass of the bob does not enter the period formula, meaning changes to bob mass have zero effect on the period. When the length of the pendulum is quadrupled (

L' = 4L

), the new period becomes

T' = 2\pi \sqrt{\frac{4L}{g}} = 2 \times \left(2\pi \sqrt{\frac{L}{g}}\right) = 2T

. Since the original period was

2.0\text{ s}

, the new period is

2 \times 2.0\text{ s} = 4.0\text{ s}

.

Step-by-Step Solution

1

Identify the formula for the period of a simple pendulum.

The period is given by

T = 2\pi \sqrt{\frac{L}{g}}

.

The period of simple harmonic motion for a simple pendulum depends only on the length of the string

L

and the acceleration due to gravity

g

, and is independent of the mass of the bob

m

.

2

Substitute the scaled values into the formula to find the new period

T'

.

T' = 2\pi \sqrt{\frac{4L}{g}} = 2 \times \left(2\pi \sqrt{\frac{L}{g}}\right) = 2T

.

Taking the square root of

4L

factors out a multiplier of

\sqrt{4} = 2

, while the change in mass from

m

to

4m

has no effect on the period.

3

Calculate the numerical value of the new period.

T' = 2 \times 2.0\text{ s} = 4.0\text{ s}

.

Multiplying the initial period of

2.0\text{ s}

by

2

yields

4.0\text{ s}

.

Key Concept

Mass Independence and Length Relationship of a Simple Pendulum

Question 10Question →

A particle undergoing simple harmonic motion moves with an angular frequency of $4\text{ rad/s}$ and an amplitude of $0.5\text{ m}$ . What is the maximum speed of the particle in $\text{m/s}$ ?

Show answer & explanation

Answer: 2

Answer

The maximum speed of the particle is

2.0\text{ m/s}

.

The magnitude of velocity in simple harmonic motion varies with displacement

x

according to

v = \omega \sqrt{A^2 - x^2}

. The speed reaches its maximum value when the particle passes through the equilibrium position (

x = 0

), giving

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

. Substituting

\omega = 4\text{ rad/s}

and

A = 0.5\text{ m}

gives

v_{\text{max}} = 4 \times 0.5 = 2.0\text{ m/s}

.

Step-by-Step Solution

1

Identify the given physical parameters.

\omega = 4\text{ rad/s}

and

A = 0.5\text{ m}

These values define the speed profile of the simple harmonic oscillator.

2

Apply the SHM formula for maximum speed.

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

Peak speed occurs at the equilibrium position where displacement is zero.

3

Substitute the values to calculate the maximum speed.

v_{\text{max}} = 4 \times 0.5 = 2.0\text{ m/s}

Multiplying angular frequency by amplitude yields the maximum linear velocity.

Key Concept

Maximum speed in Simple Harmonic Motion

Question 11Question →

What is the length of a simple pendulum that has a period of oscillation of $2.0\text{ s}$ at a location where the acceleration due to gravity is $g = \pi^2\text{ m/s}^2$ ?

0.5\text{ m}

1.0\text{ m}

2.0\text{ m}

4.0\text{ m}

Show answer & explanation

Answer: $1.0\text{ m}$

Answer

The length of the simple pendulum is

1.0\text{ m}

.

Using the period equation

T = 2\pi \sqrt{l/g}

, squaring both sides yields

T^2 = 4\pi^2 l / g

. Rearranging gives

l = \frac{T^2 g}{4\pi^2}

. Substituting

T = 2.0\text{ s}

and

g = \pi^2\text{ m/s}^2

yields

l = \frac{4 \pi^2}{4 \pi^2} = 1.0\text{ m}

.

Step-by-Step Solution

1

State the formula for the period of a simple pendulum.

T = 2\pi \sqrt{\frac{l}{g}}

This formula relates the oscillation period

T

, pendulum length

l

, and gravitational acceleration

g

.

2

Square both sides of the equation to solve for

l

.

T^2 = 4\pi^2 \left(\frac{l}{g}\right) \implies l = \frac{T^2 \cdot g}{4\pi^2}

Isolating

l

allows direct evaluation using the given numerical values.

3

Substitute

T = 2.0\text{ s}

and

g = \pi^2\text{ m/s}^2

into the expression.

l = \frac{(2.0)^2 \cdot \pi^2}{4\pi^2} = \frac{4\pi^2}{4\pi^2} = 1.0\text{ m}

Simplifying by canceling

\pi^2

and

4

gives the exact length.

Key Concept

Simple Pendulum Period and Length Relationship

Question 12Question →

A simple pendulum of length $L$ and bob mass $m$ has a period of oscillation $T$ on the surface of the Earth. The length of the pendulum is increased by $44\%$ , its bob mass is doubled to $2m$ , and the entire apparatus is transported to a planet where the acceleration due to gravity is $36\%$ less than that on Earth. What is the new period of oscillation of the pendulum?

3.0T

1.5T

1.2T

0.67T

Show answer & explanation

Answer: $1.5T$

Answer

1.5T

The period of a simple pendulum is determined by

T = 2\pi \sqrt{\frac{L}{g}}

. It is entirely independent of the mass of the bob. With a length increase of

44\%

(

L' = 1.44L

) and a gravity reduction of

36\%

(

g' = 0.64g

), the new period becomes

\sqrt{\frac{1.44}{0.64}}T = \frac{1.2}{0.8}T = 1.5T

.

Step-by-Step Solution

1

State the equation for the period of a simple pendulum

T = 2\pi \sqrt{\frac{L}{g}}

The period depends solely on length

L

and acceleration due to gravity

g

, and is independent of mass

m

.

2

Determine the new length and new acceleration due to gravity

L' = 1.44L

and

g' = 0.64g

A

44\%

increase in length yields

1 + 0.44 = 1.44

, while a

36\%

decrease in gravity yields

1 - 0.36 = 0.64

.

3

Calculate the factor of change in the period

\frac{T'}{T} = \sqrt{\frac{L'}{L} \times \frac{g}{g'}} = \sqrt{\frac{1.44}{0.64}} = \sqrt{\frac{144}{64}} = \frac{12}{8} = 1.5

Substituting the relative changes into the pendulum period ratio yields the scaling factor.

4

Express the new period in terms of

T

T' = 1.5T

Multiplying the initial period by the calculated scaling factor gives the final period.

Key Concept

Mass independence and parametric scaling of simple pendulum period in Simple Harmonic Motion

Question 13Question →

A particle of mass $0.50\text{ kg}$ executes simple harmonic motion along a straight line. When its displacement from the equilibrium position is $0.06\text{ m}$ , its speed is $0.80\text{ m/s}$ and its potential energy is $0.09\text{ J}$ . What is the magnitude of the maximum acceleration of the particle in $\text{m/s}^2$ ?

Show answer & explanation

Answer: 10

Answer

The magnitude of the maximum acceleration of the particle is

10\text{ m/s}^2

.

Using potential energy

E_p = \frac{1}{2}m\omega^2 x^2

, the angular frequency

\omega

is

10\text{ rad/s}

. Using

v = \omega\sqrt{A^2 - x^2}

, the amplitude

A

is

0.10\text{ m}

. Substituting these values into

a_{\max} = \omega^2 A

yields

10\text{ m/s}^2

.

Step-by-Step Solution

1

Find angular frequency

\omega

from potential energy.

\omega = 10\text{ rad/s}

Using potential energy

E_p = \frac{1}{2}m\omega^2 x^2

:

0.09 = \frac{1}{2}(0.50)\omega^2 (0.06)^2 \implies 0.09 = 0.0009 \omega^2 \implies \omega^2 = 100 \implies \omega = 10\text{ rad/s}

.

2

Find amplitude

A

from speed.

A = 0.10\text{ m}

Using speed

v = \omega\sqrt{A^2 - x^2}

:

0.80 = 10\sqrt{A^2 - 0.06^2} \implies 0.08 = \sqrt{A^2 - 0.0036} \implies 0.0064 = A^2 - 0.0036 \implies A^2 = 0.0100 \implies A = 0.10\text{ m}

.

3

Calculate maximum acceleration.

a_{\max} = 10\text{ m/s}^2

Using maximum acceleration formula

a_{\max} = \omega^2 A

:

a_{\max} = 100 \times 0.10 = 10\text{ m/s}^2

.

Key Concept

Simple Harmonic Motion Energy and Kinematic Relations

Question 14Question →

A $0.40\text{ kg}$ mass attached to a light helical spring undergoes simple harmonic motion on a smooth horizontal surface. If the force constant of the spring is $160\text{ N/m}$ and the amplitude of oscillation is $0.05\text{ m}$ , what is the maximum speed of the mass?

1.0\text{ m/s}

20.0\text{ m/s}

0.40\text{ m/s}

4.00\text{ m/s}

Show answer & explanation

Answer: $1.0\text{ m/s}$

Answer

The maximum speed of the mass is

1.0\text{ m/s}

.

The angular frequency of the mass-spring system is calculated using $\omega = \sqrt{k/m} = \sqrt{160/0.40} = 20\text{ rad/s}$. Multiplying this by the amplitude $A = 0.05\text{ m}$ yields a maximum speed of $v_{\text{max}} = 1.0\text{ m/s}$.

Step-by-Step Solution

1

Calculate the angular frequency ($\omega$) of the mass-spring system.

$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{160\text{ N/m}}{0.40\text{ kg}}} = \sqrt{400} = 20\text{ rad/s}$

The angular frequency of a spring-mass oscillator depends on the spring constant and the mass.

2

Determine the maximum speed ($v_{\text{max}}$) using the amplitude.

$v_{\text{max}} = \omega A = 20\text{ rad/s} \times 0.05\text{ m} = 1.0\text{ m/s}$

In simple harmonic motion, maximum speed occurs at the equilibrium position and equals the product of angular frequency and amplitude.

Key Concept

Maximum velocity in simple harmonic motion for a mass-spring system

Question 15Question →

A vertical light spring stretches by $0.10\text{ m}$ when a block is suspended from it in equilibrium. The block is then pulled down an additional $0.05\text{ m}$ from its equilibrium position and released from rest to undergo simple harmonic motion. Taking acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the speed of the block when it is at a displacement of $0.03\text{ m}$ from its equilibrium position?

0.40\text{ m/s}

0.50\text{ m/s}

0.30\text{ m/s}

0.04\text{ m/s}

Show answer & explanation

Answer: $0.40\text{ m/s}$

Answer

0.40\text{ m/s}

The system's angular frequency

\omega

is determined by the equilibrium extension

e

using

\omega = \sqrt{\frac{g}{e}} = 10\text{ rad/s}

. Combining this with the amplitude

A = 0.05\text{ m}

in the SHM speed relation

v = \omega \sqrt{A^2 - y^2}

at

y = 0.03\text{ m}

yields

v = 10 \sqrt{(0.05)^2 - (0.03)^2} = 0.40\text{ m/s}

.

Step-by-Step Solution

1

Calculate the angular frequency of the mass-spring system using static equilibrium conditions.

\omega = 10\text{ rad/s}

At equilibrium,

mg = ke \implies \frac{k}{m} = \frac{g}{e}

. Therefore,

\omega = \sqrt{\frac{g}{e}} = \sqrt{\frac{10}{0.10}} = 10\text{ rad/s}

.

2

Identify the amplitude of simple harmonic motion.

A = 0.05\text{ m}

The initial displacement from the equilibrium position when released from rest defines the amplitude of oscillation.

3

Apply the SHM velocity-displacement formula

v = \omega \sqrt{A^2 - y^2}

at

y = 0.03\text{ m}

.

v = 0.40\text{ m/s}

v = 10 \times \sqrt{(0.05)^2 - (0.03)^2} = 10 \times \sqrt{0.0025 - 0.0009} = 10 \times 0.04 = 0.40\text{ m/s}

.

Key Concept

Relating static extension to angular frequency and calculating instantaneous speed in Simple Harmonic Motion

Question 16Question →

A particle of mass $0.20\text{ kg}$ executes simple harmonic motion along a straight line. When its displacement from the equilibrium position is $0.03\text{ m}$ , its speed is $0.16\text{ m/s}$ . When its displacement is $0.04\text{ m}$ , its speed is $0.12\text{ m/s}$ . What is the total mechanical energy of the particle in millijoules ( $\text{mJ}$ )?

Show answer & explanation

Answer: 4

Answer

The total mechanical energy of the particle is

4\text{ mJ}

.

Using the relation

v^2 = \omega^2(A^2 - x^2)

for the two given state points

(0.03\text{ m}, 0.16\text{ m/s})

and

(0.04\text{ m}, 0.12\text{ m/s})

forms a set of simultaneous equations. Subtracting them yields

\omega^2 = 16\text{ rad}^2/\text{s}^2

, leading to

A^2 = 0.0025\text{ m}^2

. Substituting these values into

E = \frac{1}{2}m\omega^2 A^2

gives

E = 0.004\text{ J}

, which converts to

4\text{ mJ}

.

Step-by-Step Solution

1

Set up kinematic equations for both displacement points using

v^2 = \omega^2(A^2 - x^2)

.

0.0256 = \omega^2(A^2 - 0.0009)

and

0.0144 = \omega^2(A^2 - 0.0016)

.

The equation relates linear speed, angular frequency, amplitude, and instantaneous displacement in SHM.

2

Subtract the two simultaneous equations to eliminate

A^2

and find

\omega^2

.

0.0112 = 0.0007\omega^2 \implies \omega^2 = 16\text{ rad}^2/\text{s}^2

.

Eliminating amplitude isolates the angular frequency squared.

3

Determine

A^2

by substituting

\omega^2 = 16

back into one of the state equations.

A^2 = 0.0025\text{ m}^2 \implies A = 0.05\text{ m}

.

Amplitude is required to calculate the maximum potential or total mechanical energy.

4

Calculate total mechanical energy

E = \frac{1}{2}m\omega^2 A^2

and convert to millijoules.

E = \frac{1}{2} \times 0.20 \times 16 \times 0.0025 = 0.004\text{ J} = 4\text{ mJ}

.

Total energy in SHM is constant and proportional to mass, square of angular frequency, and square of amplitude.

Key Concept

Conservation of energy and phase-space relationship between velocity and displacement in simple harmonic motion.

Question 17Question →

An object undergoing simple harmonic motion completes $20$ complete oscillations in a time duration of $10.0\text{ s}$ . What is the period of oscillation of the object in seconds?

Show answer & explanation

Answer: 0.5

Answer

The period of oscillation is

0.5\text{ s}

.

The period of simple harmonic motion is the time taken to complete one single oscillation. Dividing the total time (

10.0\text{ s}

) by the number of oscillations (

20

) yields

0.5\text{ s}

.

Step-by-Step Solution

1

Apply the definition of oscillation period

T = \frac{t}{N}

Period

T

measures the time required for a single complete cycle.

2

Calculate the numerical value for period

T = \frac{10.0\text{ s}}{20} = 0.5\text{ s}

Dividing total elapsed time by total completed oscillations gives time per oscillation.

Key Concept

Period of Simple Harmonic Motion

Question 18Question →

A body of mass $0.50\text{ kg}$ connected to a light helical spring of force constant $32\text{ N/m}$ executes simple harmonic motion on a smooth horizontal surface. If the total mechanical energy of the oscillating system is $0.16\text{ J}$ , what is the maximum speed of the body in $\text{m/s}$ ?

Show answer & explanation

Answer: 0.8

Answer

The maximum speed of the body is

0.8\text{ m/s}

.

The total mechanical energy in simple harmonic motion is equal to the maximum kinetic energy at the equilibrium position:

E = \frac{1}{2} m v_{\text{max}}^2

. Substituting

E = 0.16\text{ J}

and

m = 0.50\text{ kg}

yields

0.16 = 0.25 v_{\text{max}}^2

, so

v_{\text{max}}^2 = 0.64

and

v_{\text{max}} = 0.8\text{ m/s}

.

Step-by-Step Solution

1

Relate total energy to maximum kinetic energy

E = \frac{1}{2} m v_{\text{max}}^2

At the equilibrium position, potential energy is zero and total mechanical energy is entirely kinetic.

2

Substitute given values into the equation

0.16 = \frac{1}{2} (0.50) v_{\text{max}}^2

Mass

m = 0.50\text{ kg}

and total energy

E = 0.16\text{ J}

are provided.

3

Solve for the maximum speed

v_{\text{max}} = \sqrt{\frac{2 \times 0.16}{0.50}} = \sqrt{0.64} = 0.8\text{ m/s}

Solving for

v_{\text{max}}

gives

0.8\text{ m/s}

.

Key Concept

Conservation of Energy in Simple Harmonic Motion

Question 19Question →

A particle executing simple harmonic motion has a maximum speed of $3.0\text{ m/s}$ and a maximum acceleration of $12.0\text{ m/s}^2$ . What is the period of oscillation of the particle?

\frac{\pi}{4}\text{ s}

\frac{\pi}{2}\text{ s}

\pi\text{ s}

2\pi\text{ s}

Show answer & explanation

Answer: $\frac{\pi}{2}\text{ s}$

Answer

The period of oscillation of the particle is

\frac{\pi}{2}\text{ s}

.

For simple harmonic motion, maximum speed is

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

and maximum acceleration is

a_{\text{max}} = \omega^2 A

. Dividing the maximum acceleration by the maximum speed gives

\omega = \frac{a_{\text{max}}}{v_{\text{max}}} = \frac{12.0}{3.0} = 4.0\text{ rad/s}

. Using the formula for the period

T = \frac{2\pi}{\omega}

, we find

T = \frac{2\pi}{4.0} = \frac{\pi}{2}\text{ s}

.

Step-by-Step Solution

1

Relate maximum speed and maximum acceleration to angular frequency

\omega = \frac{a_{\text{max}}}{v_{\text{max}}}

Since

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

and

a_{\text{max}} = \omega^2 A

, dividing

a_{\text{max}}

by

v_{\text{max}}

eliminates the amplitude

A

and gives

\omega

.

2

Calculate the angular frequency

\omega

\omega = \frac{12.0\text{ m/s}^2}{3.0\text{ m/s}} = 4.0\text{ rad/s}

Substitute the given numerical values into the expression for angular frequency.

3

Calculate the period

T

T = \frac{2\pi}{\omega} = \frac{2\pi}{4.0} = \frac{\pi}{2}\text{ s}

The period of simple harmonic motion is related to angular frequency by

T = \frac{2\pi}{\omega}

.

Key Concept

Relationship between maximum velocity, maximum acceleration, angular frequency, and period in simple harmonic motion.

Estimated Time:1m 15s

Question 20Question →

A simple pendulum has a period of oscillation of $1.6\text{ s}$ on the surface of the Earth, where the acceleration due to gravity is $10.0\text{ m/s}^2$ . What is the period of oscillation of this pendulum when placed on a moon where the acceleration due to gravity is $2.5\text{ m/s}^2$ ?

Show answer & explanation

Answer: 3.2

Answer

The period of oscillation of the pendulum on the moon is

3.2\text{ s}

.

The period of a simple pendulum is given by

T = 2\pi \sqrt{\frac{l}{g}}

. Because the length

l

is constant, period is inversely proportional to the square root of acceleration due to gravity (

T \propto \frac{1}{\sqrt{g}}

). Reducing the local gravity from

10.0\text{ m/s}^2

to

2.5\text{ m/s}^2

decreases gravity by a factor of 4, which increases the period by a factor of

\sqrt{4} = 2

. Multiplying the initial period of

1.6\text{ s}

by 2 yields

3.2\text{ s}

.

Step-by-Step Solution

1

Relate the period of oscillation of a simple pendulum to gravitational acceleration.

The period formula is

T = 2\pi \sqrt{\frac{l}{g}}

, showing that

T

is inversely proportional to

\sqrt{g}

.

The length of the pendulum

l

remains unchanged.

2

Formulate a ratio comparing the pendulum's period on the moon to its period on Earth.

\frac{T_{moon}}{T_{earth}} = \sqrt{\frac{g_{earth}}{g_{moon}}}

Dividing the two equations cancels the constant terms

2\pi

and

\sqrt{l}

.

3

Substitute the known numerical values and solve for

T_{moon}

.

T_{moon} = 1.6 \times \sqrt{\frac{10.0}{2.5}} = 1.6 \times 2.0 = 3.2\text{ s}

The ratio of gravities is 4, whose square root is 2, doubling the initial period.

Key Concept

Dependence of Simple Pendulum Period on Gravitational Acceleration

Estimated Time:1m 30s

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