Oscillations Chapter-Wise Test 16

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A particle’s displacement in SHM is \( x = 6 \sin (3t) \) (in cm). What is the maximum acceleration?

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

\( A = 6 \, \text{cm} = 0.06 \, \text{m}, \omega = 3 \, \text{s}^{-1} \).

\( a_{\text{max}} = 3^2 \times 0.06 = 9 \times 0.06 = 0.54 \, \text{m/s}^2 \).

0.36 m/s²
0.45 m/s²
0.54 m/s²
0.72 m/s²
3

A particle in SHM has \( a = -16 x \) (in SI units). What is its frequency?

For SHM, \( a = -\omega^2 x \). Given \( a = -16 x \), \( \omega^2 = 16 \Rightarrow \omega = 4 \, \text{rad/s} \).

Frequency: \( v = \frac{\omega}{2\pi} = \frac{4}{2 \times 3.14} \approx 0.637 \, \text{Hz} \).

0.5 Hz
0.637 Hz
0.7 Hz
0.8 Hz
2

A particle’s displacement is \( x = 3 \cos (2\pi t - \frac{\pi}{4}) \) (in m). What is its velocity at \( t = 0.25 \, \text{s} \)? (Take \( \sin 45^\circ = \frac{\sqrt{2}}{2} \))

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

\( A = 3 \, \text{m}, \omega = 2\pi \, \text{s}^{-1}, \phi = -\frac{\pi}{4} \).

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

\( v = -2\pi \times 3 \sin \frac{\pi}{4} = -6\pi \times \frac{\sqrt{2}}{2} \approx -13.32 \, \text{m/s} \).

-10.0 m/s
-12.0 m/s
-13.32 m/s
-15.0 m/s
3

A pendulum has \( L = 0.5 \, \text{m}, g = 9.8 \, \text{m/s}^2 \). What is its angular frequency?

\( \omega = \sqrt{\frac{g}{L}} = \sqrt{\frac{9.8}{0.5}} = \sqrt{19.6} \approx 4.43 \, \text{rad/s} \).

3.0 rad/s
4.0 rad/s
5.0 rad/s
4.43 rad/s
4

A spring-mass system oscillates with \( T = 0.6 \, \text{s} \) when \( m = 0.9 \, \text{kg} \). What is the spring constant?

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

\( 0.6 = 2\pi \sqrt{\frac{0.9}{k}} \Rightarrow \frac{0.6}{2\pi} = \sqrt{\frac{0.9}{k}} \).

\( (0.0955)^2 = \frac{0.9}{k} \Rightarrow k = \frac{0.9}{0.00912} \approx 98.68 \, \text{N/m} \).

80 N/m
90 N/m
98.68 N/m
100 N/m
3

A particle’s x-projection from circular motion is \( x = 5 \cos (4t) \) (in m). What is its maximum speed?

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

\( A = 5 \, \text{m}, \omega = 4 \, \text{s}^{-1} \).

\( v_{\text{max}} = 4 \times 5 = 20 \, \text{m/s} \).

15 m/s
18 m/s
20 m/s
22 m/s
3

A mass oscillates with \( T = 0.4 \, \text{s} \) when attached to a spring of \( k = 100 \, \text{N/m} \). What is the mass?

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

\( 0.4 = 2\pi \sqrt{\frac{m}{100}} \Rightarrow \frac{0.4}{2\pi} = \sqrt{\frac{m}{100}} \).

\( \left(\frac{0.4}{6.28}\right)^2 = \frac{m}{100} \Rightarrow m = 100 \times (0.0637)^2 \approx 0.405 \, \text{kg} \).

0.25 kg
0.405 kg
0.5 kg
0.75 kg
2

In SHM, what is true about the particle’s acceleration when its kinetic energy is at its maximum?

Kinetic energy is maximum at the mean position (\( x = 0 \)), where acceleration (\( a = -\omega^2 x \)) is zero, as the restoring force vanishes.

It is at its maximum
It is zero
It equals the velocity
It depends on the amplitude
2

Which feature of SHM explains why two particles with identical amplitude and frequency may not reach their extreme positions simultaneously?

Different phase constants (\( \phi \)) shift the oscillation cycles, causing particles to reach extremes at different times despite equal amplitude and frequency.

Variation in mass
Difference in spring constants
Difference in phase constants
Change in gravitational field
3

For a pendulum executing small oscillations, what approximation allows its motion to be treated as simple harmonic?

For small angles, \( \sin \theta \approx \theta \) (in radians), making the restoring torque (\( \tau = -mgL \theta \)) linear, a requirement for SHM.

\( \sin \theta \approx \theta \)
\( \cos \theta \approx 1 \)
\( \theta^2 \approx 0 \)
\( \tan \theta \approx \theta^2 \)
1

Performance Summary

Score:

Category Details
Total Attempts:
Total Skipped:
Total Wrong Answers:
Total Correct Answers:
Time Taken:
Average Time Taken per Question:
Accuracy:
0