Simple Harmonic Motion Practice Questions — JAMB UTME — Page 2

Question 21Question →

A particle executes simple harmonic motion along a straight line with an angular frequency of $6.0\text{ rad/s}$ . What is the magnitude of the acceleration of the particle, in $\text{m/s}^2$ , when its displacement from the mean position is $0.50\text{ m}$ ?

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Answer: 18

Answer

18.0 m/s^2

The magnitude of acceleration in simple harmonic motion is calculated using

a = \omega^2 x

. Substituting

\omega = 6.0\text{ rad/s}

and

x = 0.50\text{ m}

gives

a = (6.0)^2 \times 0.50 = 36 \times 0.50 = 18.0\text{ m/s}^2

.

Step-by-Step Solution

1

Identify the formula relating acceleration to angular frequency and displacement in SHM.

a = \omega^2 x

In simple harmonic motion, the magnitude of acceleration is directly proportional to displacement from the equilibrium position.

2

Substitute the given physical values into the equation.

a = (6.0)^2 \times 0.50

The given values are angular frequency

\omega = 6.0\text{ rad/s}

and displacement

x = 0.50\text{ m}

.

3

Compute the numerical product.

a = 18.0\text{ m/s}^2

Squaring

6.0

yields

36

, and multiplying by

0.50

gives

18.0

.

Key Concept

Acceleration in Simple Harmonic Motion

Estimated Time:1m 0s

Question 22Question →

A simple pendulum of fixed length $l$ carries a bob of mass $m$ and oscillates with a period $T$ . If the bob is replaced by another bob of mass $2m$ while maintaining the exact same string length, what is the new period of oscillation?

2T

T

\sqrt{2}T

\frac{T}{2}

Show answer & explanation

Answer: $T$

Answer

The new period of oscillation remains

T

.

The period of oscillation for a simple pendulum undergoing small displacement SHM is given by

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

. The mass of the bob

m

is not a parameter in this equation. Therefore, altering the mass while keeping length

l

constant leaves the period unchanged as

T

.

Step-by-Step Solution

1

Identify the governing equation for the period of a simple pendulum executing simple harmonic motion.

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

To analyze which physical parameters determine the period of the simple pendulum.

2

Check for mass dependence in the formula.

The mass variable

m

does not appear in

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

.

The restoring gravitational force and inertia both scale linearly with mass, causing mass to cancel out entirely.

3

Determine the period after doubling the mass at constant length.

The new period is equal to

T

.

Since length

l

and acceleration due to gravity

g

are constant, changing the mass from

m

to

2m

has no effect on period.

Key Concept

Independence of Simple Pendulum Period from Bob Mass

Question 23Question →

A particle of mass $0.20\text{ kg}$ executes simple harmonic motion with an amplitude of $0.05\text{ m}$ and a period of oscillation of $0.20\pi\text{ s}$ . What is the maximum kinetic energy of the particle in joules?

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Answer: 0.025

Answer

The maximum kinetic energy of the particle is

0.025\text{ J}

.

The maximum kinetic energy occurs at the equilibrium position where speed reaches its maximum value

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

. Substituting

\omega = \frac{2\pi}{0.20\pi} = 10\text{ rad/s}

into the maximum speed formula gives

v_{\text{max}} = 10 \times 0.05 = 0.50\text{ m/s}

. Computing kinetic energy yields

E_k = \frac{1}{2} (0.20) (0.50)^2 = 0.025\text{ J}

.

Step-by-Step Solution

1

Calculate angular frequency (\omega)

\omega = 10\text{ rad/s}

Using the relation

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

with

T = 0.20\pi\text{ s}

.

2

Calculate maximum speed (v_{max})

v_{max} = 0.50\text{ m/s}

Maximum velocity occurs at the equilibrium position and is given by

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

.

3

Calculate maximum kinetic energy (E_k)

E_k = 0.025\text{ J}

Using the kinetic energy formula

E_k = \frac{1}{2} m v^2

at maximum velocity.

Key Concept

Energy conservation and maximum speed in Simple Harmonic Motion

Question 24Question →

A body of mass $0.40\text{ kg}$ is attached to a light helical spring and set into simple harmonic motion. If the system oscillates with a period of $0.40\pi\text{ s}$ , what is the force constant of the spring in $\text{N/m}$ ?

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Answer: 10

Answer

The force constant of the spring is

10\text{ N/m}

.

The period of a mass-spring system undergoing simple harmonic motion is given by

T = 2\pi \sqrt{\frac{m}{k}}

. Substituting

m = 0.40\text{ kg}

and

T = 0.40\pi\text{ s}

yields

0.40\pi = 2\pi \sqrt{\frac{0.40}{k}}

. Dividing both sides by

2\pi

gives

0.20 = \sqrt{\frac{0.40}{k}}

. Squaring both sides yields

0.04 = \frac{0.40}{k}

, which gives

k = 10\text{ N/m}

.

Step-by-Step Solution

1

Identify the period relationship for a spring-mass SHM system

T = 2\pi \sqrt{\frac{m}{k}}

This formula relates the period of oscillation

T

to the mass

m

and spring constant

k

.

2

Substitute the given mass and period into the equation

0.40\pi = 2\pi \sqrt{\frac{0.40}{k}}

Given that

m = 0.40\text{ kg}

and

T = 0.40\pi\text{ s}

.

3

Simplify the equation by isolating the square root term

\sqrt{\frac{0.40}{k}} = 0.20

Dividing both sides of the equation by

2\pi

isolates the radical.

4

Square both sides and solve for the spring constant

k

k = 10\text{ N/m}

Squaring yields

0.04 = \frac{0.40}{k}

, which rearranges to

k = \frac{0.40}{0.04} = 10\text{ N/m}

.

Key Concept

Period of oscillation of a mass-spring system in Simple Harmonic Motion

Estimated Time:1m 30s

Question 25Question →

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

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Answer: 0.05

Answer

The displacement from the equilibrium position is

0.05\text{ m}

.

In simple harmonic motion, potential energy is

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

and kinetic energy is

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

. Setting

E_k = 3E_p

gives

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

, which simplifies to

A^2 = 4x^2

, or

x = \frac{A}{2}

. Given an amplitude

A = 0.10\text{ m}

, the displacement is

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

.

Step-by-Step Solution

1

Set up the relation between kinetic energy and potential energy using the given condition.

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

In simple harmonic motion, energy is partitioned between kinetic and potential forms based on displacement

x

.

2

Solve the algebraic equation for displacement

x

in terms of amplitude

A

.

A^2 - x^2 = 3x^2 \implies A^2 = 4x^2 \implies x = \frac{A}{2}

Canceling the common factor

\frac{1}{2}k

isolates the geometric parameters

A

and

x

.

3

Substitute the known amplitude value into the expression for

x

.

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

Plugging in

A = 0.10\text{ m}

yields the required displacement.

Key Concept

Energy Conservation in Simple Harmonic Motion

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