Oscillations Chapter-Wise Test 4

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A particle in SHM has \( x = 5 \sin (3t + \frac{\pi}{4}) \) (in m). What is its speed at \( t = 0.25 \, \text{s} \)? (Take \( \cos 45^\circ = \frac{\sqrt{2}}{2} \))

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

\( A = 5 \, \text{m}, \omega = 3 \, \text{s}^{-1}, \phi = \frac{\pi}{4} \).

At \( t = 0.25 \): \( 3 \times 0.25 + \frac{\pi}{4} = 0.75 + 0.785 \approx 1.535 \, \text{rad} \approx 88^\circ \).

\( v = 3 \times 5 \cos 88^\circ \approx 15 \times 0.0349 \approx 0.5235 \, \text{m/s} \).

0.4 m/s
0.5235 m/s
0.6 m/s
0.8 m/s
2

Two springs (\( k = 150 \, \text{N/m} \)) are attached to a \( 3 \, \text{kg} \) mass as in Fig. 13.14. What is the frequency?

Effective \( k_{\text{eff}} = 2k = 2 \times 150 = 300 \, \text{N/m} \).

\( \omega = \sqrt{\frac{k_{\text{eff}}}{m}} = \sqrt{\frac{300}{3}} = 10 \, \text{rad/s} \).

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

1.0 Hz
1.59 Hz
2.0 Hz
2.5 Hz
2

A mass oscillates with \( v = -9 \sin (3t) \) (in m/s). What is its displacement function?

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

\( \omega = 3 \, \text{s}^{-1}, v_{\text{max}} = \omega A = 9 \Rightarrow A = \frac{9}{3} = 3 \, \text{m} \).

Displacement: \( x = A \cos (\omega t) = 3 \cos (3t) \).

\( 3 \sin (3t) \)
\( 3 \cos (3t) \)
\( 9 \cos (3t) \)
\( 9 \sin (3t) \)
2

In SHM, what is the relationship between the magnitudes of maximum acceleration and maximum velocity?

Maximum velocity \( v_{\text{max}} = \omega A \), maximum acceleration \( a_{\text{max}} = \omega^2 A \). Thus, \( a_{\text{max}} = \omega v_{\text{max}} \), showing \( a_{\text{max}} \) is proportional to \( v_{\text{max}} \) via \( \omega \).

They are equal
\( a_{\text{max}} \) is proportional to \( v_{\text{max}} \)
\( v_{\text{max}} \) is the square of \( a_{\text{max}} \)
They are independent
2

Which function represents SHM? (\( \omega \) is a positive constant)

SHM requires \( a = -\omega^2 x \):

(a) \( \sin \omega t + 2 \cos 3\omega t \): Not SHM (mixed frequencies).

(b) \( 5 \cos (\omega t - \frac{\pi}{3}) \): \( a = -5\omega^2 \cos (\omega t - \frac{\pi}{3}) = -\omega^2 x \), SHM.

(c) \( e^{\omega t} \): Not periodic.

(d) \( \sin^3 \omega t \): Periodic, not SHM.

\( \sin \omega t + 2 \cos 3\omega t \)
\( 5 \cos (\omega t - \frac{\pi}{3}) \)
\( e^{\omega t} \)
\( \sin^3 \omega t \)
2

A simple pendulum has a period of \( 3 \, \text{s} \) on Earth (\( g = 9.8 \, \text{m/s}^2 \)). What is its length?

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

\( 3 = 2\pi \sqrt{\frac{L}{9.8}} \Rightarrow \sqrt{\frac{L}{9.8}} = \frac{3}{2\pi} \approx 0.477 \).

\( \frac{L}{9.8} = (0.477)^2 \Rightarrow L \approx 9.8 \times 0.227 \approx 2.23 \, \text{m} \).

1.8 m
2.0 m
2.23 m
2.5 m
3

A particle executes SHM with an amplitude of \( 5 \, \text{cm} \) and a period of \( 2 \, \text{s} \). What is the maximum speed of the particle? (Take \( \pi = 3.14 \))

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

Angular frequency: \( \omega = \frac{2\pi}{T} = \frac{2 \times 3.14}{2} = 3.14 \, \text{rad/s} \).

Amplitude: \( A = 5 \, \text{cm} = 0.05 \, \text{m} \).

\( v_{\text{max}} = 0.05 \times 3.14 = 0.157 \, \text{m/s} \).

0.12 m/s
0.157 m/s
0.20 m/s
0.25 m/s
2

What is the physical significance of the angular frequency in SHM?

Angular frequency (\( \omega = 2\pi v \)) relates to how quickly the particle oscillates, determining the rate of phase change and thus the frequency of oscillations.

It measures the maximum displacement
It determines the rate of oscillation
It indicates the restoring force magnitude
It represents the total energy
2

A spring system has \( k = 500 \, \text{N/m}, m = 2 \, \text{kg} \). What is the maximum acceleration if \( A = 8 \, \text{cm} \)?

\( \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{500}{2}} = \sqrt{250} \approx 15.81 \, \text{rad/s} \).

\( a_{\text{max}} = \omega^2 A = (15.81)^2 \times 0.08 \approx 250 \times 0.08 = 20 \, \text{m/s}^2 \).

15 m/s²
18 m/s²
20 m/s²
25 m/s²
3

A particle in SHM has \( v = -10 \sin (2t + \frac{\pi}{2}) \) (in m/s). What is its displacement function?

Velocity: \( v = -\omega A \sin (\omega t + \phi) \). Given \( \omega = 2, v_{\text{max}} = \omega A = 10 \Rightarrow A = 5 \, \text{m} \).

Comparing: \( -\omega A \sin (2t + \frac{\pi}{2}) = -10 \sin (2t + \frac{\pi}{2}) \), phase matches.

Displacement: \( x = A \cos (\omega t + \phi) = 5 \cos (2t + \frac{\pi}{2}) \).

\( 5 \sin (2t) \)
\( 5 \cos (2t + \frac{\pi}{2}) \)
\( 10 \cos (2t) \)
\( 10 \sin (2t + \frac{\pi}{2}) \)
2

A particle in SHM has an amplitude of \( 15 \, \text{cm} \) and a frequency of \( 2 \, \text{Hz} \). What is its maximum velocity? (Take \( \pi = 3.14 \))

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

\( \omega = 2\pi v = 2 \times 3.14 \times 2 = 12.56 \, \text{rad/s} \).

\( A = 0.15 \, \text{m} \).

\( v_{\text{max}} = 0.15 \times 12.56 \approx 1.884 \, \text{m/s} \).

1.5 m/s
1.8 m/s
1.884 m/s
2.0 m/s
3

Which property of an SHM system ensures that the frequency remains constant regardless of the energy imparted?

Frequency (\( v = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \)) depends solely on mass and spring constant, not energy or amplitude, due to the linear nature of the restoring force.

The linearity of the restoring force
The conservation of kinetic energy
The variation of potential energy
The amplitude dependence
1

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

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

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

At \( t = 0.5 \): \( \pi \times 0.5 + \frac{\pi}{6} = \frac{\pi}{2} + \frac{\pi}{6} = \frac{3\pi}{6} + \frac{\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3} \).

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

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

What underlies the ability of SHM to be represented as a projection of uniform circular motion?

The sinusoidal variation of displacement in SHM (\( x = A \cos (\omega t + \phi) \)) matches the projection of a circular path’s radius, reflecting identical periodic behavior.

The sinusoidal nature of the displacement
The constant angular velocity
The centripetal force equivalence
The linear velocity profile
1

A spring-mass system has \( m = 0.2 \, \text{kg}, k = 80 \, \text{N/m} \). If displaced by \( 5 \, \text{cm} \), what is the total energy?

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

\( A = 0.05 \, \text{m}, k = 80 \, \text{N/m} \).

\( E = 0.5 \times 80 \times (0.05)^2 = 0.5 \times 80 \times 0.0025 = 0.1 \, \text{J} \).

0.1 J
0.15 J
0.2 J
0.25 J
1

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