Oscillations & Waves
Q. A block on a frictionless surface is attached to a spring and undergoes simple harmonic motion. If the spring constant is 200 N/m and the mass is 2 kg, what is the period of oscillation?
A.
0.5 s
B.
1 s
C.
2 s
D.
4 s
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Solution
The period T is given by T = 2π√(m/k). Here, T = 2π√(2/200) = 2π√(0.01) = 2π(0.1) = 0.2π ≈ 0.63 s.
Correct Answer: B — 1 s
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Q. A block on a spring oscillates with a frequency of 3 Hz. What is the angular frequency of the motion?
A.
3 rad/s
B.
6 rad/s
C.
9 rad/s
D.
12 rad/s
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Solution
Angular frequency (ω) = 2πf = 2π(3) = 6 rad/s.
Correct Answer: B — 6 rad/s
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Q. A block on a spring oscillates with a period of 1.5 seconds. If the mass of the block is halved, what will be the new period?
A.
1.5 s
B.
1.22 s
C.
1.73 s
D.
1.0 s
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Solution
The period of a mass-spring system is T = 2π√(m/k). Halving the mass does not change the period since it is independent of mass in this case.
Correct Answer: A — 1.5 s
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Q. A damped harmonic oscillator has a mass of 2 kg and a damping coefficient of 0.5 kg/s. What is the damping ratio if the spring constant is 8 N/m?
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Solution
Damping ratio (ζ) = c / (2√(mk)) = 0.5 / (2√(2*8)) = 0.5 / (2√16) = 0.5 / 8 = 0.0625.
Correct Answer: B — 0.5
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Q. A damped oscillator has a time constant of 3 seconds. What is the amplitude after 6 seconds if the initial amplitude is 10 m?
A.
2.5 m
B.
5 m
C.
7.5 m
D.
10 m
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Solution
Amplitude after time t = A0 * e^(-t/τ) = 10 * e^(-6/3) = 10 * e^(-2) ≈ 2.5 m.
Correct Answer: A — 2.5 m
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Q. A damped oscillator has a time constant of 3 seconds. What is the damping coefficient if the mass is 1 kg and the spring constant is 4 N/m?
A.
1.5 kg/s
B.
2 kg/s
C.
3 kg/s
D.
4 kg/s
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Solution
Time constant (τ) = m/c, thus c = m/τ = 1/3 = 0.333 kg/s. Using c = 2ζ√(mk), we find ζ = 0.5.
Correct Answer: B — 2 kg/s
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Q. A forced oscillator has a mass of 3 kg and is driven by a force of 12 N at a frequency of 2 Hz. What is the amplitude of the oscillation if the damping coefficient is 0.1 kg/s?
A.
0.1 m
B.
0.2 m
C.
0.3 m
D.
0.4 m
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Solution
Using F = mAω², we find A = F / (mω²) = 12 / (3*(2π*2)²) ≈ 0.2 m.
Correct Answer: B — 0.2 m
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Q. A mass attached to a spring oscillates with a frequency of 2 Hz. What is the spring constant if the mass is 0.5 kg?
A.
8 N/m
B.
16 N/m
C.
32 N/m
D.
64 N/m
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Solution
Using the formula f = (1/2π)√(k/m), we can rearrange to find k = (2πf)²m. Thus, k = (2π(2))²(0.5) = 16 N/m.
Correct Answer: B — 16 N/m
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Q. A mass attached to a spring oscillates with a frequency of 3 Hz. What is the angular frequency?
A.
3 rad/s
B.
6 rad/s
C.
9 rad/s
D.
12 rad/s
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Solution
Angular frequency ω = 2πf = 2π(3) = 6 rad/s.
Correct Answer: B — 6 rad/s
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Q. A mass attached to a spring oscillates with a frequency of 3 Hz. What is the angular frequency of the motion?
A.
3 rad/s
B.
6 rad/s
C.
9 rad/s
D.
12 rad/s
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Solution
Angular frequency (ω) = 2πf = 2π(3) = 6 rad/s.
Correct Answer: B — 6 rad/s
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Q. A mass attached to a spring oscillates with a frequency of 5 Hz. What is the time period of the oscillation?
A.
0.1 s
B.
0.2 s
C.
0.5 s
D.
1.0 s
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Solution
The time period T is the reciprocal of frequency f. Thus, T = 1/f = 1/5 = 0.2 s.
Correct Answer: C — 0.5 s
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Q. A mass attached to a spring oscillates with a maximum speed of 4 m/s. If the spring constant is 100 N/m, what is the maximum displacement?
A.
0.1 m
B.
0.2 m
C.
0.4 m
D.
0.5 m
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Solution
Maximum speed v_max = ωA, where A is the amplitude. We have ω = √(k/m). Here, k = 100 N/m and m = 1 kg (assuming). Thus, A = v_max/ω = 4/10 = 0.4 m.
Correct Answer: C — 0.4 m
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Q. A mass attached to a spring oscillates with a period of 2 seconds. What is the angular frequency of the motion?
A.
0.5 rad/s
B.
1 rad/s
C.
3.14 rad/s
D.
6.28 rad/s
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Solution
Angular frequency ω = 2π/T = 2π/2 = π rad/s ≈ 3.14 rad/s.
Correct Answer: B — 1 rad/s
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Q. A mass attached to a spring oscillates with a period of 2 seconds. What is the frequency of the oscillation?
A.
0.25 Hz
B.
0.5 Hz
C.
1 Hz
D.
2 Hz
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Solution
Frequency (f) is the reciprocal of the period (T). f = 1/T = 1/2 = 0.5 Hz.
Correct Answer: B — 0.5 Hz
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Q. A mass m is attached to a spring of spring constant k. If the mass is displaced from its equilibrium position and released, what is the time period of the oscillation?
A.
2π√(m/k)
B.
2π√(k/m)
C.
π√(m/k)
D.
π√(k/m)
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Solution
The time period T of a mass-spring system in simple harmonic motion is given by T = 2π√(m/k).
Correct Answer: A — 2π√(m/k)
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Q. A mass m is attached to a spring of spring constant k. What is the angular frequency of the simple harmonic motion?
A.
√(k/m)
B.
k/m
C.
m/k
D.
1/√(km)
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Solution
The angular frequency ω is given by ω = √(k/m).
Correct Answer: A — √(k/m)
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Q. A mass on a spring oscillates with a frequency of 2 Hz. What is the angular frequency?
A.
4π rad/s
B.
2π rad/s
C.
π rad/s
D.
8π rad/s
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Solution
The angular frequency ω is given by ω = 2πf. For f = 2 Hz, ω = 2π(2) = 4π rad/s.
Correct Answer: A — 4π rad/s
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Q. A mass-spring system is subjected to a periodic force. If the amplitude of oscillation is 0.1 m and the frequency is 2 Hz, what is the maximum velocity of the mass?
A.
0.4 m/s
B.
0.2 m/s
C.
0.1 m/s
D.
0.8 m/s
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Solution
Maximum velocity (v_max) = Aω = A(2πf) = 0.1 * (2π * 2) = 0.4 m/s.
Correct Answer: A — 0.4 m/s
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Q. A mass-spring system is subjected to a periodic force. If the amplitude of the forced oscillation is 0.1 m and the damping coefficient is 0.2 kg/s, what is the maximum velocity of the oscillation?
A.
0.1 m/s
B.
0.2 m/s
C.
0.3 m/s
D.
0.4 m/s
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Solution
Maximum velocity (v_max) = Aω, where ω = 2πf. Assuming f = 1 Hz, v_max = 0.1 * 2π * 1 = 0.2 m/s.
Correct Answer: B — 0.2 m/s
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Q. A mass-spring system oscillates with a frequency of 2 Hz. If the system is damped, what is the relationship between the damped frequency and the natural frequency?
A.
Damped frequency is greater
B.
Damped frequency is equal
C.
Damped frequency is less
D.
Damped frequency is unpredictable
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Solution
In a damped system, the damped frequency is always less than the natural frequency.
Correct Answer: C — Damped frequency is less
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Q. A mass-spring system oscillates with a frequency of 2 Hz. What is the angular frequency?
A.
4π rad/s
B.
2π rad/s
C.
π rad/s
D.
8π rad/s
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Solution
The angular frequency ω is related to the frequency f by the formula ω = 2πf. Therefore, ω = 2π × 2 = 4π rad/s.
Correct Answer: A — 4π rad/s
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Q. A mass-spring system oscillates with a frequency of 2 Hz. What is the time period of the oscillation?
A.
0.5 s
B.
1 s
C.
2 s
D.
4 s
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Solution
The time period T is the reciprocal of frequency f. T = 1/f = 1/2 Hz = 0.5 s.
Correct Answer: B — 1 s
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Q. A mass-spring system oscillates with a frequency of 3 Hz. What is the angular frequency?
A.
3 rad/s
B.
6 rad/s
C.
9 rad/s
D.
12 rad/s
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Solution
Angular frequency (ω) is given by ω = 2πf. Thus, ω = 2π * 3 ≈ 6 rad/s.
Correct Answer: B — 6 rad/s
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Q. A mass-spring system oscillates with a frequency of 3 Hz. What is the angular frequency of the system?
A.
3 rad/s
B.
6 rad/s
C.
9 rad/s
D.
12 rad/s
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Solution
Angular frequency (ω) is given by ω = 2πf. Thus, ω = 2π(3) = 6 rad/s.
Correct Answer: B — 6 rad/s
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Q. A mass-spring system oscillates with a frequency of 3 Hz. What is the period of the oscillation?
A.
0.33 s
B.
0.5 s
C.
1 s
D.
2 s
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Solution
Period (T) is the reciprocal of frequency (f). T = 1/f = 1/3 = 0.33 s.
Correct Answer: A — 0.33 s
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Q. A mass-spring system oscillates with a frequency of 5 Hz. What is the period of the motion?
A.
0.2 s
B.
0.5 s
C.
1 s
D.
2 s
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Solution
Period T = 1/f = 1/5 = 0.2 s.
Correct Answer: A — 0.2 s
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Q. A mass-spring system oscillates with a natural frequency of 3 Hz. If a damping force is applied, what is the new frequency of oscillation if the damping ratio is 0.1?
A.
2.8 Hz
B.
2.9 Hz
C.
3.0 Hz
D.
3.1 Hz
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Solution
New frequency (ω_d) = ω_n√(1-ζ²) = 3√(1-0.1²) ≈ 2.9 Hz.
Correct Answer: B — 2.9 Hz
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Q. A mass-spring system oscillates with a period of 2 seconds. What is the frequency of the oscillation?
A.
0.25 Hz
B.
0.5 Hz
C.
1 Hz
D.
2 Hz
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Solution
Frequency (f) is the reciprocal of the period (T). f = 1/T = 1/2 = 0.5 Hz.
Correct Answer: B — 0.5 Hz
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Q. A mass-spring system oscillates with a period of 4 seconds. What is the frequency of the oscillation?
A.
0.25 Hz
B.
0.5 Hz
C.
1 Hz
D.
2 Hz
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Solution
Frequency f is the reciprocal of the period T. Thus, f = 1/T = 1/4 = 0.25 Hz.
Correct Answer: A — 0.25 Hz
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Q. A particle in simple harmonic motion has a maximum speed of 4 m/s and an amplitude of 2 m. What is the angular frequency?
A.
2 rad/s
B.
4 rad/s
C.
8 rad/s
D.
16 rad/s
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Solution
Maximum speed (v_max) = ωA. Thus, ω = v_max/A = 4/2 = 2 rad/s.
Correct Answer: C — 8 rad/s
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