Oscillations Chapter-Wise Test 8

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A spring-mass system oscillates with \( T = 1.0 \, \text{s} \) when \( m = 0.4 \, \text{kg} \). What is the spring constant?

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

\( 1.0 = 2\pi \sqrt{\frac{0.4}{k}} \Rightarrow \frac{1.0}{2\pi} = \sqrt{\frac{0.4}{k}} \).

\( (0.159)^2 = \frac{0.4}{k} \Rightarrow k = \frac{0.4}{0.0253} \approx 15.81 \, \text{N/m} \).

12 N/m
15 N/m
15.81 N/m
18 N/m
3

A pendulum of length \( 1.5 \, \text{m} \) oscillates with \( g = 10 \, \text{m/s}^2 \). What is its period?

\( T = 2\pi \sqrt{\frac{L}{g}} = 2\pi \sqrt{\frac{1.5}{10}} = 2\pi \sqrt{0.15} \approx 2 \times 3.14 \times 0.387 \approx 2.43 \, \text{s} \).

1.8 s
2.0 s
2.43 s
2.8 s
3

A spring system has \( m = 0.25 \, \text{kg}, k = 100 \, \text{N/m}, A = 8 \, \text{cm} \). What is the potential energy at \( x = 4 \, \text{cm} \)?

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

\( k = 100 \, \text{N/m}, x = 0.04 \, \text{m} \).

\( U = 0.5 \times 100 \times (0.04)^2 = 0.5 \times 100 \times 0.0016 = 0.08 \, \text{J} \).

0.06 J
0.08 J
0.1 J
0.12 J
2

A particle in SHM has \( a = -49 x \) (in SI units). What is its period?

For SHM, \( a = -\omega^2 x \). Given \( a = -49 x \), \( \omega^2 = 49 \Rightarrow \omega = 7 \, \text{rad/s} \).

Period: \( T = \frac{2\pi}{\omega} = \frac{2\pi}{7} \approx 0.897 \, \text{s} \).

0.7 s
0.897 s
1.0 s
1.2 s
2

A particle in SHM has \( a = -81 x \) (in SI units). What is its frequency?

For SHM, \( a = -\omega^2 x \). Given \( a = -81 x \), \( \omega^2 = 81 \Rightarrow \omega = 9 \, \text{rad/s} \).

Frequency: \( v = \frac{\omega}{2\pi} = \frac{9}{2 \times 3.14} \approx 1.43 \, \text{Hz} \).

1.0 Hz
1.43 Hz
1.5 Hz
2.0 Hz
2

A simple pendulum has a length of \( 1.44 \, \text{m} \) and oscillates with \( g = 9.8 \, \text{m/s}^2 \). What is its period?

Period: \( T = 2\pi \sqrt{\frac{L}{g}} = 2\pi \sqrt{\frac{1.44}{9.8}} \approx 2 \times 3.14 \sqrt{0.1469} \approx 2.406 \, \text{s} \).

2.0 s
2.2 s
2.406 s
2.6 s
3

A spring-mass system has \( m = 2.25 \, \text{kg}, k = 900 \, \text{N/m} \). What is its period?

Period: \( T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{2.25}{900}} = 2\pi \sqrt{0.0025} = 2\pi \times 0.05 = 0.314 \, \text{s} \).

0.2 s
0.314 s
0.4 s
0.5 s
2

A particle in SHM has a displacement \( x = 3 \cos (4t + \frac{\pi}{3}) \) (in meters). What is its acceleration at \( t = 0 \, \text{s} \)?

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

Here, \( \omega = 4 \, \text{s}^{-1}, x(0) = 3 \cos (\frac{\pi}{3}) = 3 \times 0.5 = 1.5 \, \text{m} \).

\( a(0) = -4^2 \times 1.5 = -16 \times 1.5 = -24 \, \text{m/s}^2 \).

-18 m/s²
-20 m/s²
-24 m/s²
-30 m/s²
3

A mass of \( 1.2 \, \text{kg} \) on a spring with \( k = 300 \, \text{N/m} \) has \( A = 8 \, \text{cm} \). What is the kinetic energy at \( x = 4 \, \text{cm} \)?

Total energy: \( E = \frac{1}{2} k A^2 = 0.5 \times 300 \times (0.08)^2 = 0.96 \, \text{J} \).

Potential energy: \( U = \frac{1}{2} k x^2 = 0.5 \times 300 \times (0.04)^2 = 0.24 \, \text{J} \).

Kinetic energy: \( K = E - U = 0.96 - 0.24 = 0.72 \, \text{J} \).

0.5 J
0.6 J
0.72 J
0.8 J
3

A simple pendulum has a period of \( 1.4 \, \text{s} \) on Earth (\( g = 9.8 \, \text{m/s}^2 \)). What is its length?

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

\( 1.4 = 2\pi \sqrt{\frac{L}{9.8}} \Rightarrow \sqrt{\frac{L}{9.8}} = \frac{1.4}{2\pi} \approx 0.223 \).

\( \frac{L}{9.8} = (0.223)^2 \Rightarrow L \approx 9.8 \times 0.0497 \approx 0.487 \, \text{m} \).

0.4 m
0.487 m
0.5 m
0.6 m
2

Why does the period of a simple pendulum remain constant regardless of the bob’s material?

The period \( T = 2\pi \sqrt{\frac{L}{g}} \) depends on length and gravity, not mass or material, as mass cancels out in the dynamics (force and inertia scale equally).

The amplitude compensates for material
The frequency adjusts to density
The mass cancels out in the equation
The material affects the length
3

A particle in SHM has \( x = 4 \sin (3t) \) (in m). What is its speed at \( x = 2 \, \text{m} \)?

Velocity: \( v = \pm \omega \sqrt{A^2 - x^2} \).

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

\( v = 3 \sqrt{4^2 - 2^2} = 3 \sqrt{16 - 4} = 3 \sqrt{12} = 3 \times 2\sqrt{3} \approx 10.39 \, \text{m/s} \).

8.0 m/s
10.39 m/s
12.0 m/s
14.0 m/s
2

A particle in SHM has \( x = 6 \cos (4t) \) (in m). What is its maximum acceleration?

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

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

\( a_{\text{max}} = 4^2 \times 6 = 16 \times 6 = 96 \, \text{m/s}^2 \).

80 m/s²
90 m/s²
96 m/s²
100 m/s²
3

A mass oscillates with \( v = -10 \cos (5t) \) (in m/s). What is its amplitude?

Velocity: \( v = -\omega A \sin (\omega t) \), but given \( v = -10 \cos (5t) \).

\( \omega = 5 \, \text{s}^{-1}, v_{\text{max}} = \omega A = 10 \Rightarrow A = \frac{10}{5} = 2 \, \text{m} \).

1.5 m
2.0 m
2.5 m
3.0 m
2

Which characteristic of SHM ensures that the motion repeats exactly after a fixed interval?

The sinusoidal nature of displacement (\( x = A \cos (\omega t + \phi) \)) ensures periodicity, as the cosine function repeats every \( 2\pi \), giving a fixed period \( T = \frac{2\pi}{\omega} \).

The amplitude
The phase constant
The angular frequency
The sinusoidal variation
4

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