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
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
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
Solution
Angular frequency (ω) = 2πf = 2π(3) = 6 rad/s.
Correct Answer: B — 6 rad/s
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
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
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
Solution
Angular frequency ω = 2πf = 2π(3) = 6 rad/s.
Correct Answer: B — 6 rad/s
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
Solution
Angular frequency (ω) = 2πf = 2π(3) = 6 rad/s.
Correct Answer: B — 6 rad/s
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
Solution
Frequency (f) is the reciprocal of the period (T). f = 1/T = 1/2 = 0.5 Hz.
Correct Answer: B — 0.5 Hz
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
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
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
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
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
Solution
Angular frequency (ω) is given by ω = 2πf. Thus, ω = 2π * 3 ≈ 6 rad/s.
Correct Answer: B — 6 rad/s
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
Solution
Angular frequency (ω) is given by ω = 2πf. Thus, ω = 2π(3) = 6 rad/s.
Correct Answer: B — 6 rad/s
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
Solution
Frequency (f) is the reciprocal of the period (T). f = 1/T = 1/2 = 0.5 Hz.
Correct Answer: B — 0.5 Hz
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
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
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
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
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
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
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
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
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
Solution
A pendulum swinging with a small amplitude exhibits simple harmonic motion.
Correct Answer: C — Simple harmonic motion
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
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
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
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
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
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
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
Solution
T = 2π√(m/k) = 2π√(2/200) = 1 s.
Correct Answer: B — 1 s
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
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
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
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
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
Solution
The maximum displacement in simple harmonic motion is equal to the amplitude, which is 5 cm.
Correct Answer: C — 5 cm
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
Solution
Using the wave equation v = fλ, we can find λ = v/f = 340/170 = 2 m.
Correct Answer: B — 1 m
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
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
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
Solution
Sound travels faster in water than in air, so its speed increases when it enters water.
Correct Answer: A — Increases
Q. A sound wave travels through 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.1 m
D.2 m
Solution
Using the formula v = fλ, we can rearrange to find λ = v/f. Thus, λ = 340 m/s / 1700 Hz = 0.2 m.
Correct Answer: A — 0.2 m
Q. A tuning fork produces a sound wave of frequency 440 Hz. If the speed of sound in air is 340 m/s, what is the wavelength of the sound wave?
A.0.77 m
B.0.85 m
C.0.90 m
D.1.00 m
Solution
Using the formula λ = v/f, we find λ = 340 m/s / 440 Hz = 0.7727 m, approximately 0.77 m.
Correct Answer: A — 0.77 m
Q. A tuning fork produces a sound wave of frequency 440 Hz. What is the period of this sound wave?
A.0.00227 s
B.0.00455 s
C.0.005 s
D.0.01 s
Solution
The period T is the reciprocal of frequency f. T = 1/f = 1/440 Hz = 0.00227 s.
Correct Answer: B — 0.00455 s
Q. A tuning fork produces a sound wave of frequency 440 Hz. What is the period of this wave?
A.0.00227 s
B.0.0025 s
C.0.0023 s
D.0.0020 s
Solution
The period T is the reciprocal of frequency f. T = 1/f = 1/440 Hz = 0.00227 s.
