Oscillations Chapter-Wise Test 3

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A spring-mass system oscillates with \( T = 0.9 \, \text{s} \) when \( m = 0.9 \, \text{kg} \). What is the spring constant?

\( T = 2\pi \sqrt{\frac{m}{k}} \).

\( 0.9 = 2\pi \sqrt{\frac{0.9}{k}} \Rightarrow \frac{0.9}{2\pi} = \sqrt{\frac{0.9}{k}} \).

\( (0.1432)^2 = \frac{0.9}{k} \Rightarrow k = \frac{0.9}{0.0205} \approx 43.9 \, \text{N/m} \).

40 N/m
42 N/m
43.9 N/m
45 N/m
3

A simple pendulum has a period of \( 1.8 \, \text{s} \) on Earth (\( g = 9.8 \, \text{m/s}^2 \)). What is its length?

\( T = 2\pi \sqrt{\frac{L}{g}} \).

\( 1.8 = 2\pi \sqrt{\frac{L}{9.8}} \Rightarrow \sqrt{\frac{L}{9.8}} = \frac{1.8}{2\pi} \approx 0.2865 \).

\( \frac{L}{9.8} = (0.2865)^2 \Rightarrow L \approx 9.8 \times 0.0821 \approx 0.805 \, \text{m} \).

0.7 m
0.805 m
0.9 m
1.0 m
2

A simple pendulum oscillates with a frequency of \( 0.5 \, \text{Hz} \) on Earth (\( g = 9.8 \, \text{m/s}^2 \)). What is its length?

Period: \( T = \frac{1}{v} = \frac{1}{0.5} = 2 \, \text{s} \).

\( T = 2\pi \sqrt{\frac{L}{g}} \Rightarrow 2 = 2\pi \sqrt{\frac{L}{9.8}} \).

\( \sqrt{\frac{L}{9.8}} = \frac{1}{\pi} \Rightarrow \frac{L}{9.8} = \frac{1}{\pi^2} \Rightarrow L \approx \frac{9.8}{9.87} \approx 0.99 \, \text{m} \).

0.5 m
0.99 m
1.5 m
2.0 m
2

A spring of \( k = 250 \, \text{N/m} \) has a \( 1 \, \text{kg} \) mass. If \( E = 1.25 \, \text{J} \), what is the amplitude?

Total energy: \( E = \frac{1}{2} k A^2 \).

\( 1.25 = 0.5 \times 250 \times A^2 \Rightarrow 1.25 = 125 A^2 \Rightarrow A^2 = 0.01 \Rightarrow A = 0.1 \, \text{m} \).

0.05 m
0.08 m
0.1 m
0.12 m
3

A mass oscillates with \( v = -15 \cos (3t) \) (in m/s). What is its amplitude?

Velocity: \( v = -\omega A \sin (\omega t) \), but given \( v = -15 \cos (3t) \).

\( \omega = 3 \, \text{s}^{-1}, v_{\text{max}} = \omega A = 15 \Rightarrow A = \frac{15}{3} = 5 \, \text{m} \).

3 m
4 m
5 m
6 m
3

A particle in circular motion has its x-projection as \( x = 7 \cos (\pi t) \) (in m). What is its maximum speed?

Maximum speed: \( v_{\text{max}} = \omega A \).

\( A = 7 \, \text{m}, \omega = \pi \, \text{s}^{-1} \).

\( v_{\text{max}} = \pi \times 7 \approx 3.14 \times 7 \approx 21.98 \, \text{m/s} \).

15.0 m/s
18.0 m/s
20.0 m/s
21.98 m/s
3

A particle’s x-projection from circular motion is \( x = 8 \cos (2\pi t) \) (in m). What is its maximum acceleration?

Maximum acceleration: \( a_{\text{max}} = \omega^2 A \).

\( A = 8 \, \text{m}, \omega = 2\pi \, \text{s}^{-1} \).

\( a_{\text{max}} = (2\pi)^2 \times 8 \approx 39.48 \times 8 \approx 315.84 \, \text{m/s}^2 \).

300 m/s²
310 m/s²
315.84 m/s²
320 m/s²
3

What role does the phase constant play in differentiating two SHM systems with identical amplitude and frequency?

The phase constant (\( \phi \) in \( x = A \cos (\omega t + \phi) \)) shifts the starting point of the oscillation, affecting the initial position and velocity, thus distinguishing the motions.

It determines their initial positions
It alters their maximum velocities
It changes their periods
It modifies their total energies
1

What distinguishes the restoring force in a spring-mass system from that in a simple pendulum at small amplitudes?

The spring-mass system uses elastic force (\( F = -kx \)), constant with displacement, while the pendulum’s force (\( F = -mg \sin \theta \approx -mg \theta \)) derives from gravity and varies with angle.

It is gravitational in nature
It depends on angular displacement
It varies with mass
It is elastic in origin
4

A pendulum of length \( 1.8 \, \text{m} \) oscillates with \( g = 10 \, \text{m/s}^2 \). What is its frequency?

Period: \( T = 2\pi \sqrt{\frac{L}{g}} = 2\pi \sqrt{\frac{1.8}{10}} = 2\pi \sqrt{0.18} \approx 2.68 \, \text{s} \).

Frequency: \( v = \frac{1}{T} = \frac{1}{2.68} \approx 0.373 \, \text{Hz} \).

0.3 Hz
0.35 Hz
0.373 Hz
0.4 Hz
3

A spring system has \( m = 1 \, \text{kg}, k = 100 \, \text{N/m}, A = 20 \, \text{cm} \). What is the potential energy at \( x = 0 \, \text{m} \)?

Potential energy: \( U = \frac{1}{2} k x^2 \).

At \( x = 0 \, \text{m} \), \( U = 0 \, \text{J} \).

0.5 J
1.0 J
2.0 J
0 J
3

In SHM, when does the particle experience its maximum restoring force?

The restoring force (\( F = -kx \)) is maximum at the extreme positions (\( x = \pm A \)), where displacement is greatest, coinciding with maximum potential energy.

At the mean position
Halfway to the extreme
At the extreme positions
When velocity is maximum
3

A spring of \( k = 360 \, \text{N/m} \) has a \( 1.5 \, \text{kg} \) mass. If \( E = 1.8 \, \text{J} \), what is the amplitude?

Total energy: \( E = \frac{1}{2} k A^2 \).

\( 1.8 = 0.5 \times 360 \times A^2 \Rightarrow 1.8 = 180 A^2 \Rightarrow A^2 = 0.01 \Rightarrow A = 0.1 \, \text{m} \).

0.05 m
0.08 m
0.09 m
0.1 m
4

A simple pendulum has a length of \( 0.4 \, \text{m} \) and oscillates with \( g = 9.8 \, \text{m/s}^2 \). What is its period?

Period: \( T = 2\pi \sqrt{\frac{L}{g}} = 2\pi \sqrt{\frac{0.4}{9.8}} \approx 2 \times 3.14 \sqrt{0.0408} \approx 1.27 \, \text{s} \).

1.0 s
1.27 s
1.5 s
2.0 s
2

A particle in SHM has \( x = 6 \cos (2\pi t + \frac{\pi}{4}) \) (in m). What is its speed at \( t = 0.25 \, \text{s} \)? (Take \( \sin 45^\circ = \frac{\sqrt{2}}{2} \))

Velocity: \( v = -\omega A \sin (\omega t + \phi) \).

\( A = 6 \, \text{m}, \omega = 2\pi \, \text{s}^{-1}, \phi = \frac{\pi}{4} \).

At \( t = 0.25 \): \( 2\pi \times 0.25 + \frac{\pi}{4} = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4} \).

\( v = -2\pi \times 6 \sin \frac{3\pi}{4} = -12\pi \times \frac{\sqrt{2}}{2} \approx -26.64 \, \text{m/s} \).

-24 m/s
-25 m/s
-26 m/s
-26.64 m/s
4

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