Oscillations & Waves
Q. A pendulum swings back and forth with a period of 1 second. If the length of the pendulum is doubled, what will be the new period?
A.
1 s
B.
1.41 s
C.
2 s
D.
4 s
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Solution
The period of a simple pendulum is given by T = 2π√(L/g). If L is doubled, T becomes T' = 2π√(2L/g) = √2 * T ≈ 1.41 s.
Correct Answer: B — 1.41 s
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Q. A pendulum swings with a maximum angle of 30 degrees. What is the approximate period of the pendulum if its length is 1 m?
A.
1.0 s
B.
1.5 s
C.
2.0 s
D.
2.5 s
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Solution
The period T of a simple pendulum is given by T = 2π√(L/g). Here, L = 1 m and g = 9.8 m/s². Thus, T = 2π√(1/9.8) ≈ 2.0 s.
Correct Answer: C — 2.0 s
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Q. A pendulum swings with a period of 1 second. If the length of the pendulum is increased to four times its original length, what will be the new period?
A.
1 s
B.
2 s
C.
4 s
D.
√4 s
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Solution
The period of a pendulum is given by T = 2π√(L/g). If L is increased to 4L, T becomes 2π√(4L/g) = 2T = 2 seconds.
Correct Answer: B — 2 s
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Q. A pendulum swings with a period of 1 second. If the length of the pendulum is increased by a factor of 4, what will be the new period?
A.
1 s
B.
2 s
C.
4 s
D.
√4 s
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Solution
The period T = 2π√(L/g). If L is increased by a factor of 4, T increases by a factor of √4 = 2.
Correct Answer: B — 2 s
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Q. A pendulum swings with a period of 1 second. If the length of the pendulum is tripled, what will be the new period?
A.
1 s
B.
2 s
C.
3 s
D.
√3 s
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Solution
The period T = 2π√(L/g). If L is tripled, T becomes √3 times longer, so T = 2 s.
Correct Answer: B — 2 s
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Q. A pendulum swings with a period of 1 second. What is the length of the pendulum?
A.
0.25 m
B.
0.5 m
C.
1 m
D.
2 m
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Solution
Length L = (gT^2)/(4π^2) = (9.8 * 1^2)/(4 * π^2) ≈ 1 m.
Correct Answer: C — 1 m
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Q. A pendulum swings with a period of 1.5 seconds. What is the angular frequency of the pendulum?
A.
2π/1.5 rad/s
B.
4π/3 rad/s
C.
π/1.5 rad/s
D.
3π/2 rad/s
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Solution
Angular frequency (ω) = 2π/T = 2π/1.5 = 4π/3 rad/s.
Correct Answer: B — 4π/3 rad/s
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Q. A pendulum swings with a period of 2 seconds. What is the length of the pendulum?
A.
0.5 m
B.
1 m
C.
2 m
D.
4 m
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Solution
The period T of a simple pendulum is given by T = 2π√(L/g). Rearranging gives L = (T²g)/(4π²). Using g = 9.8 m/s², L = (2² * 9.8)/(4π²) ≈ 1 m.
Correct Answer: B — 1 m
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Q. A pendulum swings with a period T. What is the period of a pendulum of length 4L?
A.
2T
B.
T/2
C.
T√2
D.
2√2T
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Solution
The period T of a simple pendulum is given by T = 2π√(L/g). For length 4L, T' = 2π√(4L/g) = 2T.
Correct Answer: A — 2T
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Q. A pendulum swings with a small amplitude. The restoring force acting on the pendulum is proportional to which of the following?
A.
Displacement from equilibrium
B.
Velocity
C.
Acceleration
D.
Mass
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Solution
In simple harmonic motion, the restoring force is proportional to the displacement from the equilibrium position.
Correct Answer: A — Displacement from equilibrium
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Q. A pendulum swings with a small amplitude. What type of motion does it exhibit?
A.
Linear motion
B.
Rotational motion
C.
Simple harmonic motion
D.
Circular motion
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Solution
A pendulum swinging with a small amplitude exhibits simple harmonic motion.
Correct Answer: C — Simple harmonic motion
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Q. A simple harmonic oscillator has a frequency of 5 Hz. What is the time period of the oscillator?
A.
0.2 s
B.
0.5 s
C.
1 s
D.
2 s
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Solution
The time period (T) is the reciprocal of frequency (f). T = 1/f = 1/5 = 0.2 s.
Correct Answer: A — 0.2 s
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Q. A simple harmonic oscillator has a mass of 2 kg and a spring constant of 200 N/m. What is the angular frequency of the oscillator?
A.
5 rad/s
B.
10 rad/s
C.
20 rad/s
D.
15 rad/s
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Solution
The angular frequency ω is given by the formula ω = √(k/m). Here, k = 200 N/m and m = 2 kg. Thus, ω = √(200/2) = √100 = 10 rad/s.
Correct Answer: B — 10 rad/s
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Q. A simple harmonic oscillator has a mass of 2 kg and a spring constant of 50 N/m. What is the angular frequency of the oscillator?
A.
5 rad/s
B.
10 rad/s
C.
15 rad/s
D.
20 rad/s
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Solution
The angular frequency ω is given by the formula ω = √(k/m). Here, k = 50 N/m and m = 2 kg. Thus, ω = √(50/2) = √25 = 5 rad/s.
Correct Answer: B — 10 rad/s
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Q. A simple harmonic oscillator has a maximum displacement of 0.1 m and a maximum speed of 2 m/s. What is the angular frequency?
A.
10 rad/s
B.
20 rad/s
C.
5 rad/s
D.
15 rad/s
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Solution
ω = V_max/A = 2/0.1 = 20 rad/s.
Correct Answer: A — 10 rad/s
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Q. A simple harmonic oscillator has a maximum displacement of 0.1 m. What is the maximum potential energy if the spring constant is 200 N/m?
A.
1 J
B.
2 J
C.
3 J
D.
4 J
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Solution
Maximum potential energy (PE) = (1/2)kA^2 = (1/2)(200)(0.1^2) = 1 J.
Correct Answer: B — 2 J
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Q. A simple harmonic oscillator has a spring constant of 200 N/m and a mass of 2 kg. What is its period?
A.
0.5 s
B.
1 s
C.
2 s
D.
4 s
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Solution
T = 2π√(m/k) = 2π√(2/200) = 1 s.
Correct Answer: B — 1 s
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Q. A simple harmonic oscillator has a total energy E. If the amplitude is halved, what will be the new total energy?
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Solution
The total energy E is proportional to the square of the amplitude. If the amplitude is halved, the new energy will be E/4.
Correct Answer: A — E/4
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Q. A simple harmonic oscillator has a total energy of 50 J and an amplitude of 10 cm. What is the spring constant?
A.
200 N/m
B.
500 N/m
C.
1000 N/m
D.
2000 N/m
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Solution
Total energy E = (1/2)kA^2. 50 = (1/2)k(0.1^2) => k = 500 N/m.
Correct Answer: B — 500 N/m
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Q. A simple harmonic oscillator has a total energy of 50 J. If the amplitude is doubled, what will be the new total energy?
A.
50 J
B.
100 J
C.
200 J
D.
400 J
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Solution
Total energy in SHM is proportional to the square of the amplitude. If amplitude is doubled, energy increases by a factor of 4.
Correct Answer: C — 200 J
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Q. A simple harmonic oscillator has an amplitude A and a maximum speed v_max. What is the relationship between v_max and A?
A.
v_max = Aω
B.
v_max = A/ω
C.
v_max = A²ω
D.
v_max = A/2ω
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Solution
The maximum speed v_max of a simple harmonic oscillator is given by v_max = Aω, where ω is the angular frequency.
Correct Answer: A — v_max = Aω
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Q. A simple harmonic oscillator has an amplitude of 5 cm. What is the maximum displacement from the mean position?
A.
0 cm
B.
2.5 cm
C.
5 cm
D.
10 cm
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Solution
The maximum displacement in simple harmonic motion is equal to the amplitude, which is 5 cm.
Correct Answer: C — 5 cm
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Q. A sound wave travels at 340 m/s. If its frequency is 170 Hz, what is its wavelength?
A.
0.5 m
B.
1 m
C.
2 m
D.
3 m
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Solution
Using the wave equation v = fλ, we can find λ = v/f = 340/170 = 2 m.
Correct Answer: B — 1 m
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Q. A sound wave travels at a speed of 340 m/s in air. If its frequency is 1700 Hz, what is its wavelength?
A.
0.2 m
B.
0.5 m
C.
1.0 m
D.
2.0 m
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Solution
Using the wave equation v = fλ, we can find the wavelength: λ = v/f = 340 m/s / 1700 Hz = 0.2 m.
Correct Answer: B — 0.5 m
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Q. A sound wave travels at a speed of 340 m/s. If its frequency is 170 Hz, what is its wavelength?
A.
0.5 m
B.
1 m
C.
2 m
D.
3 m
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Solution
Using the wave equation v = fλ, we can find the wavelength: λ = v/f = 340/170 = 2 m.
Correct Answer: B — 1 m
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Q. A sound wave travels from air into water. What happens to its speed?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Becomes zero
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Solution
Sound travels faster in water than in air, so its speed increases when it enters water.
Correct Answer: A — Increases
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Q. A sound wave travels in air at a speed of 340 m/s. If the frequency of the sound is 1700 Hz, what is the wavelength?
A.
0.2 m
B.
0.5 m
C.
2 m
D.
1 m
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Solution
Wavelength λ = v/f = 340 m/s / 1700 Hz = 0.2 m.
Correct Answer: A — 0.2 m
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Q. A sound wave travels in air at a speed of 340 m/s. If the frequency of the sound wave is 1700 Hz, what is its wavelength?
A.
0.2 m
B.
0.5 m
C.
2 m
D.
1 m
Show solution
Solution
Wavelength λ = v/f = 340 m/s / 1700 Hz = 0.2 m.
Correct Answer: B — 0.5 m
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Q. A sound wave travels in air at a speed of 340 m/s. If the frequency of the sound is 170 Hz, what is the wavelength?
A.
2 m
B.
1 m
C.
0.5 m
D.
0.2 m
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Solution
Wavelength λ = v/f = 340 m/s / 170 Hz = 2 m.
Correct Answer: A — 2 m
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Q. A sound wave travels through a medium with a speed of 340 m/s and has a frequency of 1700 Hz. What is its wavelength?
A.
0.2 m
B.
0.5 m
C.
2 m
D.
1 m
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Solution
Wavelength = Speed / Frequency = 340 m/s / 1700 Hz = 0.2 m.
Correct Answer: B — 0.5 m
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