1. Home
    2. >
  2. Neet Physics Class 11 Mock Test Papers
    3. >
  3. Physics Chapter Wise Test Papers
    4. >
  4. Oscillations Chapter-Wise Test 3

Oscillations Chapter-Wise Test 3

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A pendulum has \( L = 1.96 \, \text{m}, g = 9.8 \, \text{m/s}^2 \). What is its angular frequency?

\( \omega = \sqrt{\frac{g}{L}} = \sqrt{\frac{9.8}{1.96}} = \sqrt{5} \approx 2.24 \, \text{rad/s} \).

2.0 rad/s
2.24 rad/s
2.5 rad/s
3.0 rad/s
2

A particle in SHM follows \( x = 2 \sin (4t + \frac{\pi}{2}) \) (in m). What is its acceleration at \( t = 0 \, \text{s} \)?

Acceleration: \( a = -\omega^2 x \).

\( \omega = 4 \, \text{s}^{-1}, x(0) = 2 \sin (\frac{\pi}{2}) = 2 \times 1 = 2 \, \text{m} \).

\( a = -4^2 \times 2 = -16 \times 2 = -32 \, \text{m/s}^2 \).

-24 m/s²
-28 m/s²
-32 m/s²
-36 m/s²
3

A mass of \( 2.0 \, \text{kg} \) on a spring with \( k = 800 \, \text{N/m} \) has \( A = 5 \, \text{cm} \). What is the kinetic energy at \( x = 2.5 \, \text{cm} \)?

Total energy: \( E = \frac{1}{2} k A^2 = 0.5 \times 800 \times (0.05)^2 = 1 \, \text{J} \).

Potential energy: \( U = \frac{1}{2} k x^2 = 0.5 \times 800 \times (0.025)^2 = 0.25 \, \text{J} \).

Kinetic energy: \( K = E - U = 1 - 0.25 = 0.75 \, \text{J} \).

0.5 J
0.75 J
0.8 J
1.0 J
2

A spring system has \( m = 1.0 \, \text{kg}, k = 400 \, \text{N/m}, A = 6 \, \text{cm} \). What is the potential energy at \( x = 3 \, \text{cm} \)?

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

\( k = 400 \, \text{N/m}, x = 0.03 \, \text{m} \).

\( U = 0.5 \times 400 \times (0.03)^2 = 0.5 \times 400 \times 0.0009 = 0.18 \, \text{J} \).

0.12 J
0.15 J
0.18 J
0.2 J
3

A spring-mass system has \( m = 0.25 \, \text{kg} \) and \( k = 100 \, \text{N/m} \). What is its period of oscillation?

Period: \( T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{0.25}{100}} = 2\pi \sqrt{0.0025} = 2\pi \times 0.05 \approx 0.314 \, \text{s} \).

0.2 s
0.314 s
0.4 s
0.5 s
2

A simple pendulum has a length of \( 2.25 \, \text{m} \) and oscillates with \( g = 9.8 \, \text{m/s}^2 \). What is its period?

Period: \( T = 2\pi \sqrt{\frac{L}{g}} = 2\pi \sqrt{\frac{2.25}{9.8}} \approx 2 \times 3.14 \sqrt{0.2296} \approx 3.01 \, \text{s} \).

2.5 s
2.8 s
3.01 s
3.5 s
3

A particle in SHM has \( a = -64 x \) (in SI units). What is its period?

For SHM, \( a = -\omega^2 x \). Given \( a = -64 x \), \( \omega^2 = 64 \Rightarrow \omega = 8 \, \text{rad/s} \).

Period: \( T = \frac{2\pi}{\omega} = \frac{2\pi}{8} = \frac{\pi}{4} \approx 0.785 \, \text{s} \).

0.6 s
0.785 s
0.9 s
1.0 s
2

A particle’s displacement is \( x = 5 \sin (2\pi t + \frac{\pi}{6}) \) (in m). What is its velocity at \( t = 0.5 \, \text{s} \)? (Take \( \sin 60^\circ = \frac{\sqrt{3}}{2} \))

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

\( A = 5 \, \text{m}, \omega = 2\pi \, \text{s}^{-1}, \phi = \frac{\pi}{6} \).

At \( t = 0.5 \): \( 2\pi \times 0.5 + \frac{\pi}{6} = \pi + \frac{\pi}{6} = \frac{7\pi}{6} \).

\( v = 2\pi \times 5 \cos \frac{7\pi}{6} = 10\pi \cos (180^\circ - 30^\circ) = 10\pi \left(-\frac{\sqrt{3}}{2}\right) \approx -27.14 \, \text{m/s} \).

-25 m/s
-26 m/s
-27.14 m/s
-28 m/s
3

A mass of \( 2 \, \text{kg} \) on a spring with \( k = 200 \, \text{N/m} \) has \( A = 10 \, \text{cm} \). What is the kinetic energy at \( x = 5 \, \text{cm} \)?

Total energy: \( E = \frac{1}{2} k A^2 = 0.5 \times 200 \times (0.1)^2 = 1 \, \text{J} \).

Potential energy: \( U = \frac{1}{2} k x^2 = 0.5 \times 200 \times (0.05)^2 = 0.25 \, \text{J} \).

Kinetic energy: \( K = E - U = 1 - 0.25 = 0.75 \, \text{J} \).

0.5 J
0.6 J
0.75 J
0.9 J
3

A particle in SHM has \( x = 3 \cos (2\pi t + \frac{\pi}{3}) \) (in m). What is its speed at \( t = 0.5 \, \text{s} \)? (Take \( \sin 120^\circ = \frac{\sqrt{3}}{2} \))

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

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

At \( t = 0.5 \): \( 2\pi \times 0.5 + \frac{\pi}{3} = \pi + \frac{\pi}{3} = \frac{4\pi}{3} \).

\( v = -2\pi \times 3 \sin \frac{4\pi}{3} = -6\pi \sin (180^\circ - 60^\circ) = -6\pi \left(-\frac{\sqrt{3}}{2}\right) \approx 16.31 \, \text{m/s} \).

14.0 m/s
15.0 m/s
16.31 m/s
18.0 m/s
3

Performance Summary

Score:

Category Details
Total Attempts:
Total Skipped:
Total Wrong Answers:
Total Correct Answers:
Time Taken:
Average Time Taken per Question:
Accuracy:
0
error: Content is protected !!