Oscillations Chapter-Wise Test 2

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A particle in SHM has \( x = 5 \sin (2t) \) (in m). What is its speed at \( x = 2.5 \, \text{m} \)?

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

\( A = 5 \, \text{m}, \omega = 2 \, \text{s}^{-1}, x = 2.5 \, \text{m} \).

\( v = 2 \sqrt{5^2 - 2.5^2} = 2 \sqrt{25 - 6.25} = 2 \sqrt{18.75} \approx 8.66 \, \text{m/s} \).

7.0 m/s
8.66 m/s
9.0 m/s
10.0 m/s
2

A mass of \( 3 \, \text{kg} \) is attached to a spring with \( k = 300 \, \text{N/m} \). If displaced by \( 15 \, \text{cm} \), what is the total energy?

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

\( A = 0.15 \, \text{m}, k = 300 \, \text{N/m} \).

\( E = \frac{1}{2} \times 300 \times (0.15)^2 = 0.5 \times 300 \times 0.0225 = 3.375 \, \text{J} \).

3.375 J
2.5 J
4.0 J
5.0 J
1

A pendulum of length \( 0.8 \, \text{m} \) oscillates with \( g = 10 \, \text{m/s}^2 \). What is its frequency?

Period: \( T = 2\pi \sqrt{\frac{L}{g}} = 2\pi \sqrt{\frac{0.8}{10}} = 2\pi \sqrt{0.08} \approx 1.78 \, \text{s} \).

Frequency: \( v = \frac{1}{T} = \frac{1}{1.78} \approx 0.562 \, \text{Hz} \).

0.4 Hz
0.5 Hz
0.562 Hz
0.7 Hz
3

A pendulum has \( L = 0.64 \, \text{m}, g = 9.8 \, \text{m/s}^2 \). What is its angular frequency?

\( \omega = \sqrt{\frac{g}{L}} = \sqrt{\frac{9.8}{0.64}} \approx \sqrt{15.31} \approx 3.91 \, \text{rad/s} \).

3.5 rad/s
3.7 rad/s
3.9 rad/s
3.91 rad/s
4

Two identical springs (\( k = 70 \, \text{N/m} \)) are attached to a \( 1.4 \, \text{kg} \) mass as in Fig. 13.14. What is the frequency?

Effective \( k_{\text{eff}} = 2k = 2 \times 70 = 140 \, \text{N/m} \).

\( \omega = \sqrt{\frac{k_{\text{eff}}}{m}} = \sqrt{\frac{140}{1.4}} = \sqrt{100} = 10 \, \text{rad/s} \).

\( v = \frac{\omega}{2\pi} = \frac{10}{2 \times 3.14} \approx 1.59 \, \text{Hz} \).

1.0 Hz
1.5 Hz
1.59 Hz
2.0 Hz
3

Two identical springs (\( k = 60 \, \text{N/m} \)) are attached to a \( 1.2 \, \text{kg} \) mass as in Fig. 13.14. What is the frequency?

Effective \( k_{\text{eff}} = 2k = 2 \times 60 = 120 \, \text{N/m} \).

\( \omega = \sqrt{\frac{k_{\text{eff}}}{m}} = \sqrt{\frac{120}{1.2}} = \sqrt{100} = 10 \, \text{rad/s} \).

\( v = \frac{\omega}{2\pi} = \frac{10}{2 \times 3.14} \approx 1.59 \, \text{Hz} \).

1.0 Hz
1.59 Hz
2.0 Hz
2.5 Hz
2

A particle in SHM has \( x = 5 \cos (2t + \frac{\pi}{4}) \) (in m). What is its acceleration at \( t = 0 \, \text{s} \)? (Take \( \cos 45^\circ = \frac{\sqrt{2}}{2} \))

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

\( \omega = 2 \, \text{s}^{-1}, x(0) = 5 \cos \frac{\pi}{4} = 5 \times \frac{\sqrt{2}}{2} \approx 3.536 \, \text{m} \).

\( a = -2^2 \times 3.536 = -4 \times 3.536 \approx -14.14 \, \text{m/s}^2 \).

-12 m/s²
-13 m/s²
-14.14 m/s²
-15 m/s²
3

A particle in SHM has an amplitude of \( 10 \, \text{cm} \) and a frequency of \( 2.5 \, \text{Hz} \). What is its maximum velocity? (Take \( \pi = 3.14 \))

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

\( \omega = 2\pi v = 2 \times 3.14 \times 2.5 = 15.7 \, \text{rad/s} \).

\( A = 0.1 \, \text{m} \).

\( v_{\text{max}} = 0.1 \times 15.7 = 1.57 \, \text{m/s} \).

1.57 m/s
1.8 m/s
2.0 m/s
2.5 m/s
1

Two identical springs (\( k = 100 \, \text{N/m} \)) are attached to a \( 1 \, \text{kg} \) mass as in Fig. 13.14. What is the period?

Net force: \( F = -2kx \), so effective \( k_{\text{eff}} = 2k = 200 \, \text{N/m} \).

\( T = 2\pi \sqrt{\frac{m}{k_{\text{eff}}}} = 2\pi \sqrt{\frac{1}{200}} \approx 0.44 \, \text{s} \).

0.3 s
0.5 s
0.6 s
0.44 s
4

A simple pendulum has a length of \( 0.9 \, \text{m} \) and oscillates with \( g = 9.8 \, \text{m/s}^2 \). What is its period?

Period: \( T = 2\pi \sqrt{\frac{L}{g}} = 2\pi \sqrt{\frac{0.9}{9.8}} \approx 2 \times 3.14 \sqrt{0.0918} \approx 1.9 \, \text{s} \).

1.5 s
1.9 s
2.2 s
2.5 s
2

In an ideal SHM system, what happens to the kinetic energy as the particle approaches the extreme position?

Kinetic energy decreases as the particle approaches the extreme position, where velocity becomes zero, and potential energy reaches its maximum.

It increases
It remains constant
It decreases
It becomes equal to the total energy
3

The projection of a particle in circular motion on the x-axis is \( x = 3 \cos (2\pi t) \) (in m). What is the radius of the circular path?

For SHM as projection of circular motion, radius = amplitude.

Here, \( x = A \cos (\omega t) \), so \( A = 3 \, \text{m} \).

Radius = \( 3 \, \text{m} \).

1.5 m
2.0 m
3.0 m
4.0 m
3

A particle in SHM has \( x = 2 \sin (4t + \frac{\pi}{3}) \) (in m). What is its acceleration at \( t = 0 \, \text{s} \)? (Take \( \cos \frac{\pi}{3} = 0.5 \))

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

\( \omega = 4 \, \text{s}^{-1}, x(0) = 2 \sin \frac{\pi}{3} = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3} \approx 1.732 \, \text{m} \).

\( a = -4^2 \times 1.732 = -16 \times 1.732 \approx -27.71 \, \text{m/s}^2 \).

-20 m/s²
-25 m/s²
-27.71 m/s²
-30 m/s²
3

What property of SHM allows its displacement to be expressed as a linear combination of sine and cosine functions?

The harmonic nature of SHM, driven by a linear restoring force, results in sinusoidal motion, which can be written as \( x = A \sin (\omega t) + B \cos (\omega t) \), a general solution to the SHM differential equation.

The amplitude variation
The phase constant shift
The angular frequency
The harmonic nature of motion
4

A spring-mass system has \( m = 0.5 \, \text{kg}, k = 200 \, \text{N/m} \). If the amplitude is \( 4 \, \text{cm} \), what is the total energy?

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

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

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

0.1 J
0.16 J
0.2 J
0.25 J
2

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