Oscillations Chapter-Wise Test 5

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A particle in SHM has \( x = 6 \sin (2\pi t + \frac{\pi}{4}) \) (in cm). What is its acceleration at \( t = 0.25 \, \text{s} \)?

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

\( \omega = 2\pi \, \text{s}^{-1}, x = 0.06 \cos (2\pi \times 0.25 + \frac{\pi}{4}) = 0.06 \cos (\frac{\pi}{2} + \frac{\pi}{4}) = 0.06 \cos \frac{3\pi}{4} \).

\( \cos \frac{3\pi}{4} = -\frac{\sqrt{2}}{2} \), so \( x = -0.06 \frac{\sqrt{2}}{2} \approx -0.0424 \, \text{m} \).

\( a = -(2\pi)^2 \times (-0.0424) \approx 39.48 \times 0.0424 \approx 1.67 \, \text{m/s}^2 \).

1.0 m/s²
1.5 m/s²
1.67 m/s²
2.0 m/s²
3

Why does the frequency of a spring-mass system increase when a stiffer spring is used?

Frequency \( v = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \) increases with a larger spring constant (\( k \)), as a stiffer spring (higher \( k \)) provides a stronger restoring force, speeding up oscillations.

The mass decreases
The amplitude reduces
The period increases
The restoring force strengthens
4

A particle in SHM has \( x = 3 \cos (2t + \frac{\pi}{6}) \) (in m). What is its speed at \( t = 0.5 \, \text{s} \)? (Take \( \sin \frac{4\pi}{6} = 1 \))

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

\( A = 3 \, \text{m}, \omega = 2 \, \text{s}^{-1}, \phi = \frac{\pi}{6} \).

At \( t = 0.5 \): \( 2 \times 0.5 + \frac{\pi}{6} = 1 + \frac{\pi}{6} = \frac{\pi}{3} + \frac{\pi}{6} = \frac{\pi}{2} \).

\( v = -2 \times 3 \sin (\frac{\pi}{2}) = -6 \times 1 = -6 \, \text{m/s} \).

-4.0 m/s
-5.0 m/s
-7.0 m/s
-6.0 m/s
4

A simple pendulum has a frequency of \( 0.3 \, \text{Hz} \) on Earth (\( g = 9.8 \, \text{m/s}^2 \)). What is its length?

Period: \( T = \frac{1}{v} = \frac{1}{0.3} \approx 3.333 \, \text{s} \).

\( T = 2\pi \sqrt{\frac{L}{g}} \Rightarrow 3.333 = 2\pi \sqrt{\frac{L}{9.8}} \).

\( \sqrt{\frac{L}{9.8}} = \frac{3.333}{2\pi} \approx 0.531 \Rightarrow \frac{L}{9.8} = (0.531)^2 \Rightarrow L \approx 2.76 \, \text{m} \).

2.5 m
2.76 m
3.0 m
3.5 m
2

In a simple pendulum executing SHM, what ensures that the period is independent of the initial angular displacement?

For small angles, the approximation \( \sin \theta \approx \theta \) holds, making the period \( T = 2\pi \sqrt{\frac{L}{g}} \) independent of amplitude, a feature of linear systems.

The small angle approximation
The mass of the bob
The gravitational field strength
The length variation
1

A particle in SHM has an amplitude of \( 12 \, \text{cm} \) and a period of \( 1.2 \, \text{s} \). What is its maximum velocity? (Take \( \pi = 3.14 \))

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

\( \omega = \frac{2\pi}{T} = \frac{2 \times 3.14}{1.2} \approx 5.23 \, \text{rad/s} \).

\( A = 0.12 \, \text{m} \).

\( v_{\text{max}} = 0.12 \times 5.23 \approx 0.628 \, \text{m/s} \).

0.5 m/s
0.628 m/s
0.75 m/s
1.0 m/s
2

A spring-mass system has \( m = 1.5 \, \text{kg}, k = 600 \, \text{N/m} \). What is its period?

Period: \( T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{1.5}{600}} = 2\pi \sqrt{0.0025} = 2\pi \times 0.05 \approx 0.314 \, \text{s} \).

0.2 s
0.314 s
0.4 s
0.5 s
2

A particle’s displacement is \( x = 7 \sin (2t + \frac{\pi}{4}) \) (in m). What is its velocity at \( t = 0 \, \text{s} \)? (Take \( \cos 45^\circ = \frac{\sqrt{2}}{2} \))

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

\( A = 7 \, \text{m}, \omega = 2 \, \text{s}^{-1}, \phi = \frac{\pi}{4} \).

At \( t = 0 \): \( v = 2 \times 7 \cos \frac{\pi}{4} = 14 \times \frac{\sqrt{2}}{2} = 7\sqrt{2} \approx 9.9 \, \text{m/s} \).

8.0 m/s
9.0 m/s
9.9 m/s
10.5 m/s
3

A mass of \( 2 \, \text{kg} \) is attached to a spring with \( k = 200 \, \text{N/m} \). What is the frequency of oscillation?

Angular frequency: \( \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{200}{2}} = 10 \, \text{rad/s} \).

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

1.0 Hz
1.59 Hz
2.0 Hz
2.5 Hz
2

A spring-mass system has \( T = 0.2 \, \text{s}, m = 0.8 \, \text{kg} \). What is the spring constant?

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

\( 0.2 = 2\pi \sqrt{\frac{0.8}{k}} \Rightarrow \frac{0.2}{2\pi} = \sqrt{\frac{0.8}{k}} \).

\( (0.0318)^2 = \frac{0.8}{k} \Rightarrow k = \frac{0.8}{0.00101} \approx 792 \, \text{N/m} \).

600 N/m
700 N/m
792 N/m
800 N/m
3

Why does the period of a spring-mass system differ fundamentally from that of a simple pendulum in terms of gravitational influence?

The spring-mass period (\( T = 2\pi \sqrt{\frac{m}{k}} \)) lacks gravitational dependence, relying on elasticity, while the pendulum’s period (\( T = 2\pi \sqrt{\frac{L}{g}} \)) varies with gravity.

The spring adjusts to gravitational changes
The mass cancels gravitational effects
The amplitude compensates for gravity
The restoring force excludes gravity
4

In SHM, why does the particle’s velocity lead its displacement by \( \pi/2 \) radians?

Displacement (\( x = A \cos (\omega t + \phi) \)) and velocity (\( v = -\omega A \sin (\omega t + \phi) \)) differ by \( \pi/2 \) radians because the cosine and sine functions are shifted by this phase, reflecting their derivative relationship.

Velocity depends on amplitude
Acceleration is constant
Velocity is the derivative of displacement
Displacement is maximum when velocity is zero
3

A particle in SHM has \( x = 3 \sin (4t + \frac{\pi}{3}) \) (in m). What is its acceleration at \( t = 0.25 \, \text{s} \)? (Take \( \cos 60^\circ = 0.5 \))

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

\( \omega = 4 \, \text{s}^{-1}, x = 3 \sin (4 \times 0.25 + \frac{\pi}{3}) = 3 \sin (1 + \frac{\pi}{3}) \approx 3 \sin 1.571 \approx 3 \, \text{m} \).

\( a = -4^2 \times 3 = -16 \times 3 = -48 \, \text{m/s}^2 \).

-40 m/s²
-45 m/s²
-48 m/s²
-50 m/s²
3

Which of the following functions represents SHM? (Assume \( \omega \) is a positive constant)

For SHM, acceleration \( a = -\omega^2 x \). Check by differentiating twice:

(a) \( x = \sin \omega t + \cos 2\omega t \): Not SHM (different frequencies).

(b) \( x = 2 \sin (\omega t + \frac{\pi}{4}) \): \( v = 2\omega \cos (\omega t + \frac{\pi}{4}), a = -2\omega^2 \sin (\omega t + \frac{\pi}{4}) = -\omega^2 x \). SHM.

(c) \( x = e^{-\omega t} \): Not periodic, not SHM.

(d) \( x = \log (\omega t) \): Not periodic, not SHM.

\( \sin \omega t + \cos 2\omega t \)
\( 2 \sin (\omega t + \frac{\pi}{4}) \)
\( e^{-\omega t} \)
\( \log (\omega t) \)
2

Two springs, each of \( k = 50 \, \text{N/m} \), are connected in parallel to a \( 2 \, \text{kg} \) mass. What is the period of oscillation?

Effective spring constant in parallel: \( k_{\text{eff}} = k_1 + k_2 = 50 + 50 = 100 \, \text{N/m} \).

Period: \( T = 2\pi \sqrt{\frac{m}{k_{\text{eff}}}} = 2\pi \sqrt{\frac{2}{100}} = 2\pi \sqrt{0.02} \approx 0.89 \, \text{s} \) (using \( \pi \approx 3.14 \)).

0.63 s
0.89 s
1.0 s
1.26 s
2

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