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  4. Oscillations Chapter-Wise Test 2

Oscillations Chapter-Wise Test 2

Correct answer Carries: 4.

Wrong Answer Carries: -1.

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

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

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

\( v_{\text{max}} = \pi \times 8 \approx 3.14 \times 8 \approx 25.12 \, \text{m/s} \).

20 m/s
22 m/s
25.12 m/s
28 m/s
3

A particle in SHM has \( x = 4 \cos (2t - \frac{\pi}{6}) \) (in m). What is its speed at \( t = 0.5 \, \text{s} \)? (Take \( \sin 30^\circ = 0.5 \))

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

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

At \( t = 0.5 \): \( 2 \times 0.5 - \frac{\pi}{6} = 1 - \frac{\pi}{6} \approx 0.476 \, \text{rad} \approx 27.3^\circ \).

\( v = -2 \times 4 \sin (27.3^\circ) \approx -8 \times 0.46 \approx -3.68 \, \text{m/s} \).

-3.0 m/s
-3.68 m/s
-4.0 m/s
-5.0 m/s
2

A spring of \( k = 450 \, \text{N/m} \) has a \( 1.5 \, \text{kg} \) mass. If \( E = 2.25 \, \text{J} \), what is the amplitude?

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

\( 2.25 = 0.5 \times 450 \times A^2 \Rightarrow 2.25 = 225 A^2 \Rightarrow A^2 = 0.01 \Rightarrow A = 0.1 \, \text{m} \).

0.05 m
0.08 m
0.09 m
0.1 m
4

A simple pendulum has a period of \( 1 \, \text{s} \) on Earth (\( g = 9.8 \, \text{m/s}^2 \)). What is its period on the Moon (\( g = 1.63 \, \text{m/s}^2 \))?

\( T \propto \frac{1}{\sqrt{g}} \).

\( \frac{T_{\text{Moon}}}{T_{\text{Earth}}} = \sqrt{\frac{g_{\text{Earth}}}{g_{\text{Moon}}}} = \sqrt{\frac{9.8}{1.63}} \approx \sqrt{6} \approx 2.45 \).

\( T_{\text{Moon}} = 1 \times 2.45 \approx 2.45 \, \text{s} \).

1.5 s
2.0 s
2.45 s
3.0 s
3

For a motion to be simple harmonic, the restoring force must satisfy which condition when plotted against displacement?

In SHM, the restoring force is proportional to displacement and opposite in direction (\( F = -kx \)). When plotted, this yields a straight line through the origin with a negative slope.

It forms a straight line with a negative slope
It forms a parabola opening upward
It is a constant value regardless of displacement
It forms a sinusoidal curve
1

A particle in SHM has \( x = 4 \sin (4t + \frac{\pi}{6}) \) (in m). What is its speed at \( t = 0.25 \, \text{s} \)? (Take \( \cos 60^\circ = 0.5 \))

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

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

At \( t = 0.25 \): \( 4 \times 0.25 + \frac{\pi}{6} = 1 + \frac{\pi}{6} \approx 1.523 \, \text{rad} \approx 87^\circ \).

\( v = 4 \times 4 \cos 87^\circ \approx 16 \times 0.052 \approx 0.832 \, \text{m/s} \).

0.6 m/s
0.8 m/s
0.832 m/s
0.9 m/s
3

A spring system has \( m = 0.5 \, \text{kg}, k = 200 \, \text{N/m}, A = 8 \, \text{cm} \). What is the potential energy at \( x = 4 \, \text{cm} \)?

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

\( k = 200 \, \text{N/m}, x = 0.04 \, \text{m} \).

\( U = 0.5 \times 200 \times (0.04)^2 = 0.5 \times 200 \times 0.0016 = 0.16 \, \text{J} \).

0.12 J
0.14 J
0.16 J
0.18 J
3

A particle in SHM has \( a = -25 x \) (in SI units). What is its period?

For SHM, \( a = -\omega^2 x \). Given \( a = -25 x \), \( \omega^2 = 25 \Rightarrow \omega = 5 \, \text{rad/s} \).

Period: \( T = \frac{2\pi}{\omega} = \frac{2\pi}{5} \approx 1.256 \, \text{s} \).

1.0 s
1.256 s
1.5 s
2.0 s
2

A body oscillates with SHM according to \( x = 4 \cos (2\pi t + \frac{\pi}{6}) \) (in SI units). What is its velocity at \( t = 0.5 \, \text{s} \)? (Take \( \sin \frac{4\pi}{3} = -\frac{\sqrt{3}}{2} \))

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

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

At \( t = 0.5 \, \text{s} \): \( \omega t + \phi = 2\pi \times 0.5 + \frac{\pi}{6} = \pi + \frac{\pi}{6} = \frac{7\pi}{6} \).

\( \sin \frac{7\pi}{6} = \sin (180^\circ + 30^\circ) = -\sin 30^\circ = -\frac{1}{2} \).

\( v = -2\pi \times 4 \times \left(-\frac{1}{2}\right) = 4\pi \, \text{m/s} \approx 12.56 \, \text{m/s} \).

8.0 m/s
10.0 m/s
12.56 m/s
14.0 m/s
3

Why does the period of a spring-mass system remain unaffected by changes in gravitational field strength?

The period \( T = 2\pi \sqrt{\frac{m}{k}} \) depends only on mass and spring constant, not gravity, which affects pendulums but not spring systems.

The spring constant adjusts to gravity
The mass cancels the gravitational effect
The amplitude compensates for gravity
The restoring force is independent of gravity
4

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