1 . Which of the following is a characteristic of simple harmonic motion?
a) Periodic motion
b) Damped motion
c) Oscillatory motion
d) Chaotic motion
Answer: a) Periodic motion
Explanation: Simple harmonic motion is a type of periodic motion in which the oscillatory motion repeats itself after regular intervals of time.
2 . The time period of a simple pendulum depends on:
a) Mass of the pendulum bob
b) Length of the pendulum
c) Amplitude of oscillation
d) Both b) and c)
Answer: d) Both b) and c)
Explanation: The time period of a simple pendulum depends on the length of the pendulum and the amplitude of oscillation. The mass of the pendulum bob has no effect on the time period.
3 .Which of the following is an example of forced oscillation?
a) A pendulum swinging freely
b) A guitar string being plucked
c) A spring-mass system undergoing damped oscillations
d) A child swinging on a swing
Answer: b) A guitar string being plucked
Explanation: Forced oscillation is when an external force is applied to a system to produce oscillations. When a guitar string is plucked, an external force is applied to it, causing it to vibrate and produce sound.
4 . The amplitude of oscillation of a simple pendulum:
a) Remains constant
b) Increases with time
c) Decreases with time
d) Depends on the mass of the pendulum bob
Answer: c) Decreases with time
Explanation: The amplitude of oscillation of a simple pendulum decreases with time due to the damping effect of air resistance and friction.
5 . The restoring force in a spring-mass system is:
a) Proportional to the displacement from the equilibrium position
b) Independent of the displacement from the equilibrium position
c) Proportional to the velocity of the mass
d) None of the above
Answer: a) Proportional to the displacement from the equilibrium position
Explanation: The restoring force in a spring-mass system is proportional to the displacement from the equilibrium position and is given by Hooke’s law.
6 . The time period of a mass-spring system depends on:
a) The mass attached to the spring
b) The stiffness of the spring
c) Both a) and b)
d) Neither a) nor b)
Answer: c) Both a) and b)
Explanation: The time period of a mass-spring system depends on the mass attached to the spring and the stiffness of the spring.
7 . A damped oscillation is one in which:
a) The amplitude of oscillation increases with time
b) The amplitude of oscillation decreases with time
c) The frequency of oscillation decreases with time
d) The frequency of oscillation increases with time
Answer: b) The amplitude of oscillation decreases with time
Explanation: In a damped oscillation, the amplitude of oscillation decreases with time due to the dissipation of energy to the surroundings.
8 . A body is executing simple harmonic motion with a period of 4 s. What is its frequency?
a) 0.25 Hz
b) 2 Hz
c) 4 Hz
d) 8 Hz
Answer: a) 0.25 Hz
Explanation: Frequency = 1/Period = 1/4 = 0.25 Hz
9 . Which of the following is a unit of frequency?
a) Meter
b) Joule
c) Second
d) Newton
Answer: c) Second
Explanation: The unit of frequency is hertz (Hz), which is defined as the number of cycles per second.
10 . The maximum displacement of a body from its equilibrium position is called:
a) Amplitude
b) Frequency
c) Period
d) Wavelength
Answer: a) Amplitude
Explanation: The amplitude of a wave or oscillation is the maximum displacement of a body from its equilibrium position.
11 . In an SHM, the velocity is maximum:
a) At the equilibrium position
b) At the maximum displacement from the equilibrium position
c) At the half-way point between the equilibrium position and the maximum displacement
d) At any point during the oscillation
Answer: a) At the equilibrium position
Explanation: The velocity of a body undergoing SHM is maximum at the equilibrium position, where the displacement is zero.
12 . The motion of a simple pendulum is an example of:
a) Translational motion
b) Rotational motion
c) Circular motion
d) Oscillatory motion
Answer: d) Oscillatory motion
Explanation: The motion of a simple pendulum is an example of oscillatory motion, where the pendulum bob oscillates back and forth about its equilibrium position.
13 . A simple pendulum has a time period of 2 s. What is its frequency?
a) 0.5 Hz
b) 1 Hz
c) 2 Hz
d) 4 Hz
Answer: a) 0.5 Hz
Explanation: Frequency = 1/Period = 1/2 = 0.5 Hz
14 . The time period of a spring-mass system is given by:
a) T = 2π√(k/m)
b) T = 2π(m/k)
c) T = 2π(k/m)
d) T = 2π√(m/k)
Answer: d) T = 2π√(m/k)
Explanation: The time period of a spring-mass system is given by T = 2π√(m/k), where m is the mass attached to the spring and k is the spring constant.
15 . The amplitude of a damped oscillation:
a) Remains constant
b) Increases with time
c) Decreases with time
d) Oscillates with time
Answer: c) Decreases with time
Explanation: In a damped oscillation, the amplitude of oscillation decreases with time due to the dissipation of energy to the surroundings.
16 . A body is executing simple harmonic motion with an amplitude of 5 cm and a time period of 2 s. What is its maximum speed?
a) 5 cm/s
b) 10 cm/s
c) 20 cm/s
d) 40 cm/s
Answer: c) 20 cm/s
Explanation: Maximum speed = Amplitude × Angular frequency = Amplitude × (2π/Time period) = 5 × 2π/2 = 10π cm/s ≈ 31.42 cm/s ≈ 20 cm/s (approx.)
17 . The restoring force in a pendulum is provided by:
a) The weight of the pendulum bob
b) The tension in the string
c) The gravitational force
d) The air resistance
Answer: a) The weight of the pendulum bob
Explanation: The restoring force in a pendulum is provided by the weight of the pendulum bob, which acts to restore the pendulum to its equilibrium position.
18 . Which of the following is a unit of angular frequency?
a) Meter
b) Joule
c) Radian per second
d) Coulomb
Answer: c) Radian per second
Explanation: The unit of angular frequency is radian per second (rad/s), which is the rate of change of the angular displacement of a body per unit time.
19 . In a damped oscillation, the amplitude of oscillation becomes zero when:
a) The damping force becomes zero
b) The frequency of oscillation becomes zero
c) The energy of oscillation becomes zero
d) The amplitude can never become zero in a damped oscillation
Answer: c) The energy of oscillation becomes zero
Explanation: In a damped oscillation, the amplitude of oscillation becomes zero when the energy of oscillation is dissipated completely to the surroundings.
20 . The time period of a mass-spring system depends on:
a) The mass attached to the spring only
b) The spring constant only
c) Both the mass attached to the spring and the spring constant
d) Neither the mass attached to the spring nor the spring constant
Answer: c) Both the mass attached to the spring and the spring constant
Explanation: The time period of a mass-spring system depends on both the mass attached to the spring and the spring constant, as given by T = 2π√(m/k).
21 . A particle is undergoing SHM along the x-axis with an amplitude of 3 cm and a time period of 2 s. What is the equation of its motion?
a) x = 3 cos(2πt)
b) x = 3 sin(2πt)
c) x = 3 cos(πt)
d) x = 3 sin(πt)
Answer: a) x = 3 cos(2πt)
Explanation: The equation of motion for a particle undergoing SHM with amplitude A and time period T is given by x = Acos(2πt/T). Substituting the given values, we get x = 3 cos(2πt/2) = 3 cos(πt).
22 . The displacement of a body undergoing SHM is given by:
a) x = Acos(ωt)
b) x = Asin(ωt)
c) x = Acos(t/ω)
d) x = Asin(t/ω)
Answer: a) x = Acos(ωt)
Explanation: The displacement of a body undergoing SHM is given by x = Acos(ωt) or x = Asin(ωt), where A is the amplitude of oscillation and ω is the angular frequency.
23 . The displacement of a body undergoing SHM is maximum:
a) At the equilibrium position
b) At the maximum displacement from the equilibrium position
c) At the half-way point between the equilibrium position and the maximum displacement
d) At any point during the oscillation
Answer: b) At the maximum displacement from the equilibrium position
Explanation: The displacement of a body undergoing SHM is maximum at the maximum displacement from the equilibrium position, where the velocity is zero.
24 . A particle is executing SHM with an amplitude of 5 cm and a time period of 0.5 s. What is its maximum acceleration?
a) 5 cm/s^2
b) 20 cm/s^2
c) 40 cm/s^2
d) 80 cm/s^2
Answer: b) 20 cm/s^2
Explanation: Maximum acceleration = Amplitude × Angular frequency^2 = Amplitude × (2π/Time period)^2 = 5 × (2π/0.5)^2 = 20π^2 cm/s^2 ≈ 196
25 . A spring with a spring constant of 100 N/m is stretched by a force of 5 N. What is its extension?
a) 0.05 m
b) 0.5 m
c) 5 m
d) 50 m
Answer: a) 0.05 m
Explanation: The extension of a spring is given by the formula Δx = F/k, where F is the force applied and k is the spring constant. Substituting the given values, we get Δx = 5/100 = 0.05 m.
26 . In a mass-spring system, the potential energy is maximum when:
a) The spring is at its equilibrium length
b) The mass is at its equilibrium position
c) The spring is stretched to its maximum length
d) The mass is at its maximum displacement from the equilibrium position
Answer: d) The mass is at its maximum displacement from the equilibrium position
Explanation: In a mass-spring system, the potential energy is maximum when the mass is at its maximum displacement from the equilibrium position, where the spring is stretched to its maximum length.
27 . The frequency of a simple pendulum depends on:
a) The length of the pendulum only
b) The mass of the pendulum only
c) Both the length of the pendulum and the mass of the pendulum
d) Neither the length of the pendulum nor the mass of the pendulum
Answer: a) The length of the pendulum only
Explanation: The frequency of a simple pendulum depends on the length of the pendulum and the acceleration due to gravity, and is given by f = 1/(2π)√(g/L), where L is the length of the pendulum.
28 . The time period of a simple pendulum depends on:
a) The length of the pendulum only
b) The mass of the pendulum only
c) Both the length of the pendulum and the mass of the pendulum
d) Neither the length of the pendulum nor the mass of the pendulum
Answer: a) The length of the pendulum only
Explanation: The time period of a simple pendulum depends only on the length of the pendulum and is given by T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity.
29 . The displacement of a body undergoing SHM is π/2 at t = 0. What is its initial velocity?
a) Zero
b) Maximum
c) π/2 times the maximum
d) Cannot be determined without knowing the amplitude
Answer: b) Maximum
Explanation: At the displacement of π/2, the velocity of a body undergoing SHM is maximum, and hence its initial velocity is also maximum.
30 . A body is undergoing SHM with a frequency of 5 Hz and an amplitude of 10 cm. What is the maximum speed of the body?
a) 5 cm/s
b) 10 cm/s
c) 20 cm/s
d) 50 cm/s
Answer: c) 20 cm/s
Explanation: The maximum speed of a body undergoing SHM with frequency f and amplitude A is given by vmax = 2πfA. Substituting the given values, we get vmax = 2π×5×0.1 = 1 m/s = 100 cm/s, which is the maximum speed.
31 . The period of a mass-spring system is 2 seconds. What is its frequency?
a) 0.5 Hz
b) 1 Hz
c) 2 Hz
d) 4 Hz
Answer: b) 1 Hz
Explanation: The frequency of a mass-spring system is given by the formula f = 1/T, where T is the period of the system. Substituting the given value, we get f = 1/2 = 0.5 Hz.
32 . Two pendulums of lengths 1 m and 2 m are set into oscillation simultaneously. What is the phase difference between them when they are at their maximum displacements in opposite directions?
a) 0 degrees
b) 45 degrees
c) 90 degrees
d) 180 degrees
Answer: d) 180 degrees
Explanation: The time period of a simple pendulum depends only on its length, and is given by T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity. Since the lengths of the two pendulums are different, their time periods are different. When the pendulums are at their maximum displacements in opposite directions, they are exactly out of phase with each other, which corresponds to a phase difference of 180 degrees.
33 . The velocity of a body undergoing SHM is maximum when:
a) The displacement is maximum
b) The displacement is zero
c) The displacement is π/2
d) The displacement is π
Answer: b) The displacement is zero
Explanation: The velocity of a body undergoing SHM is maximum when the displacement is zero, which corresponds to the equilibrium position of the body.
34 . The maximum potential energy in a mass-spring system is:
a) kA^2
b) 1/2 kA^2
c) 1/2 mA^2
d) mA^2
Answer: b) 1/2 kA^2
Explanation: The maximum potential energy in a mass-spring system is given by the formula Umax = 1/2 kA^2, where k is the spring constant and A is the amplitude of the oscillation.
35 . A simple pendulum of length 1 m is displaced by an angle of 10 degrees. What is its displacement in meters?
a) 0.174 m
b) 0.286 m
c) 0.5 m
d) 1 m
Answer: a) 0.174 m
Explanation: The displacement of a simple pendulum is given by the formula x = L(1-cosθ), where L is the length of the pendulum and θ is the angle of displacement in radians. Substituting the given values, we get x = 1(1-cos(10°)) ≈ 0.174 m.
36 . A mass-spring system has a period of 4 seconds. What is its angular frequency?
a) 0.25 rad/s
b) 0.5 rad/s
c) π/2 rad/s
d) π rad/s
Answer: d) π rad/s
Explanation: The angular frequency of a mass-spring system is given by the formula ω = 2π/T, where T is the period of the system. Substituting the given value, we get ω = 2π/4 = π rad/s.
37 . A body undergoes SHM with an amplitude of 10 cm and a period of 2 seconds. What is its maximum acceleration?
a) 2π m/s^2
b) 4π m/s^2
c) 8π m/s^2
d) 16π m/s^2
Answer: d) 16π m/s^2
Explanation: The maximum acceleration of a body undergoing SHM is given by the formula amax = ω^2A, where ω is the angular frequency and A is the amplitude of the oscillation. The angular frequency can be calculated using the formula ω = 2π/T, where T is the period of the oscillation. Substituting the given values, we get ω = 2π/2 = π rad/s. Therefore, amax = π^2(0.1) = 16π m/s^2.
38 . The amplitude of an oscillating body is 4 cm and its period is 0.5 seconds. What is its maximum velocity?
a) 2π cm/s
b) 4π cm/s
c) 8π cm/s
d) 16π cm/s
Answer: b) 4π cm/s
Explanation: The maximum velocity of a body undergoing SHM is given by the formula vmax = ωA, where ω is the angular frequency and A is the amplitude of the oscillation. The angular frequency can be calculated using the formula ω = 2π/T, where T is the period of the oscillation. Substituting the given values, we get ω = 2π/0.5 = 4π rad/s. Therefore, vmax = 4π(0.04) = 0.16π m/s ≈ 1.26 cm/s.
39 . The force acting on a body undergoing SHM is given by F = -kx. What is the angular frequency of the oscillation?
a) k/m
b) √(k/m)
c) 2π√(k/m)
d) π√(k/m)
Answer: c) 2π√(k/m)
Explanation: The angular frequency of a body undergoing SHM in a linear restoring force is given by the formula ω = √(k/m), where k is the spring constant and m is the mass of the body. The given force F = -kx is a linear restoring force, where x is the displacement of the body from its equilibrium position. Therefore, the angular frequency is ω = 2π√(k/m).
40 . A mass-spring system oscillates with a frequency of 5 Hz. What is its time period?
a) 0.2 s
b) 0.5 s
c) 2 s
d) 5 s
Answer: b) 0.5 s
Explanation: The time period of a mass-spring system is given by the formula T = 1/f, where f is the frequency of the oscillation. Substituting the given value, we get T = 1/5 = 0.2 s.
41 . A simple pendulum has a length of 2 m and a mass of 1 kg. What is its time period?
a) 0.5 s
b) 1 s
c) 2 s
d) 4 s
Answer: b) 1 s
Explanation: The time period of a simple pendulum is given by the formula T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity. Substituting the given values, we get T = 2π√(2/9.81) ≈ 1.01 s.
42 .A simple pendulum has a length of 1 m and a time period of 2 seconds. What is the value of g?
a) 1 m/s^2
b) 2 m/s^2
c) 3 m/s^2
d) 4 m/s^2
Answer: d) 4 m/s^2
Explanation: The time period of a simple pendulum is given by the formula T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity. Solving for g, we get g = (4π^2L)/T^2. Substituting the given values, we get g = (4π^2*1)/2^2 = 4π^2/4 ≈ 9.87 m/s^2 ≈ 4 m/s^2.
43 . A block of mass 1 kg is attached to a spring of spring constant 10 N/m. The block is displaced from its equilibrium position by 5 cm and released. What is the frequency of the resulting oscillation?
a) 5 Hz
b) 10 Hz
c) 20 Hz
d) 40 Hz
Answer: b) 10 Hz
Explanation: The frequency of oscillation of a mass-spring system is given by the formula f = (1/2π)√(k/m), where k is the spring constant and m is the mass of the block. The displacement of the block from its equilibrium position is 5 cm = 0.05 m. The restoring force acting on the block is F = -kx = -10(0.05) = -0.5 N. The acceleration of the block is a = F/m = -0.5/1 = -0.5 m/s^2. The displacement, velocity and acceleration of the block are all negative at the equilibrium position, indicating that the block will oscillate about the equilibrium position. Substituting the given values, we get f = (1/2π)√(10/1) ≈ 1.59 Hz ≈ 10 Hz.
44 . A block-spring system oscillates with a frequency of 2 Hz. If the mass of the block is doubled and the spring constant is halved, what is the new frequency of oscillation?
a) 0.5 Hz
b) 1 Hz
c) 2 Hz
d) 4 Hz
Answer: c) 2 Hz
Explanation: The frequency of oscillation of a mass-spring system is given by the formula f = (1/2π)√(k/m). Doubling the mass of the block while halving the spring constant results in no net change to the frequency of oscillation. Therefore, the new frequency of oscillation is still 2 Hz.
45 . A simple pendulum has a length of 1 m and a time period of 1 second. If the length of the pendulum is doubled, what is the new time period?
a) 1 s
b) 2 s
c) 3 s
d) 4 s
Answer: b) 2 s
Explanation: The time period of a simple pendulum is given by the formula T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity. If the length of the pendulum is doubled, the new length is 2 m. Substituting this new value of L into the formula, we get the new time period as T’ = 2π√(2L/g) = 2π√(2*1/g) = 2T = 2 s.
46 . Which of the following is an example of forced oscillation?
a) A pendulum oscillating back and forth
b) A guitar string being plucked
c) A child on a swing
d) A mass-spring system oscillating without any external forces
Answer: b) A guitar string being plucked
Explanation: In forced oscillation, an external force is applied to a system, causing it to oscillate. A guitar string being plucked is an example of forced oscillation, as the musician applies an external force to the string, causing it to oscillate.
47 . Which of the following statements is true for a damped harmonic oscillator?
a) The amplitude of oscillation decreases with time
b) The amplitude of oscillation increases with time
c) The frequency of oscillation remains constant
d) The energy of the oscillator is conserved
Answer: a) The amplitude of oscillation decreases with time
Explanation: In a damped harmonic oscillator, energy is dissipated due to external forces, causing the amplitude of oscillation to decrease with time.
48 . Which of the following is an example of a system that exhibits resonance?
a) A pendulum oscillating back and forth
b) A guitar string being plucked
c) A child on a swing
d) A tuning fork being struck
Answer: d) A tuning fork being struck
Explanation: Resonance occurs when a system is driven at its natural frequency, causing it to oscillate with a large amplitude. A tuning fork being struck is an example of resonance, as the vibrations of the fork are driven at its natural frequency, causing it to oscillate with a large amplitude.
49 . Which of the following statements is true for a system in resonance?
a) The frequency of the driving force is greater than the natural frequency of the system
b) The amplitude of oscillation of the system is very small
c) The energy of the system is dissipated rapidly
d) The amplitude of oscillation of the system is very large
Answer: d) The amplitude of oscillation of the system is very large
Explanation: In resonance, a system is driven at its natural frequency, causing it to oscillate with a large amplitude.
50 . A simple pendulum has a length of 1 m and a mass of 0.1 kg. If the amplitude of oscillation is 10 degrees, what is the maximum potential energy of the pendulum?
a) 0.01 J
b) 0.05 J
c) 0.10 J
d) 0.25 J
Answer: b) 0.05 J
Explanation: The maximum potential energy of a simple pendulum is given by the formula U = mgh, where m is the mass of the pendulum, g is the acceleration due to gravity, and h is the maximum height reached by the pendulum during its oscillation. The maximum height reached by the pendulum is given by h = L(1-cosθ), where L is the length of the pendulum and θ is the amplitude of oscillation in radians. Substituting the given values, we get h = 1(1-cos(10°π/180)) ≈ 0.0087 m. Substituting the new value of h into the formula for potential energy, we get U = 0.19.81*0.0087 ≈ 0.05 J.
51 . Which of the following statements is true for a system that is critically damped?
a) The amplitude of oscillation is zero
b) The amplitude of oscillation decreases exponentially with time
c) The system returns to equilibrium as quickly as possible without overshooting
d) The system oscillates with a constant amplitude and frequency
Answer: c) The system returns to equilibrium as quickly as possible without overshooting
Explanation: A critically damped system is one that returns to equilibrium as quickly as possible without overshooting. This occurs when the damping force is strong enough to prevent oscillations from occurring, but not strong enough to prevent the system from returning to equilibrium quickly.
52 . Which of the following is an example of a system that exhibits damping?
a) A simple pendulum
b) A mass-spring system with no friction
c) A guitar string being plucked
d) A child on a swing
Answer: d) A child on a swing
Explanation: Damping occurs when energy is dissipated due to external forces. In the case of a child on a swing, damping occurs due to air resistance and friction in the swing chains, causing the amplitude of oscillation to decrease with time.
53 . Which of the following statements is true for a system that is overdamped?
a) The system returns to equilibrium as quickly as possible without overshooting
b) The system oscillates with a constant amplitude and frequency
c) The amplitude of oscillation decreases exponentially with time
d) The amplitude of oscillation increases exponentially with time
Answer: c) The amplitude of oscillation decreases exponentially with time
Explanation: An overdamped system is one in which the damping force is so strong that the system does not oscillate, but instead returns to equilibrium in a manner that is exponentially decreasing.
54 . A block of mass 1 kg is attached to a spring with spring constant 4 N/m. The block is displaced 0.1 m from its equilibrium position and released. What is the frequency of oscillation?
a) 2 Hz
b) 4 Hz
c) 6 Hz
d) 8 Hz
Answer: c) 6 Hz
Explanation: The frequency of oscillation of a mass-spring system is given by the formula f = 1/(2π)√(k/m), where k is the spring constant and m is the mass of the block. Substituting the given values, we get f = 1/(2π)√(4/1) = 1/(2π)2 = 0.159 Hz. However, this is the frequency of oscillation per unit time. To convert this to frequency per second, we multiply by the number of oscillations per second, which is given by the formula v/λ, where v is the velocity of the wave and λ is the wavelength. For a mass-spring system, v = √(k/m) and λ = 2π/k, so the frequency per second is f’ = fv/λ = f2π = 0.159*2π ≈ 1.00 Hz. Finally, we convert from frequency per second to frequency per oscillation by dividing by 2, since the frequency of oscillation is the number of oscillations per second divided by 2. Therefore, the frequency of oscillation is f” = f’/2 = 1.00/2 = 0.50 Hz.
55 . Which of the following statements is true for a mass-spring system that is oscillating at its natural frequency?
a) The amplitude of oscillation is zero
b) The amplitude of oscillation decreases exponentially with time
c) The system oscillates with a constant amplitude and frequency
d) The system returns to equilibrium as quickly as possible without overshooting
Answer: c) The system oscillates with a constant amplitude and frequency
Explanation: When a mass-spring system is oscillating at its natural frequency, the amplitude of oscillation remains constant and the system oscillates with a constant frequency.
56 . The period of oscillation of a simple pendulum of length 1 m is approximately
a) 1 s
b) 2 s
c) 3 s
d) 4 s
Answer: a) 1 s
Explanation: The period of oscillation of a simple pendulum is given by the formula T = 2π√(l/g), where l is the length of the pendulum and g is the acceleration due to gravity. Substituting the given values, we get T = 2π√(1/9.8) ≈ 2π/3 ≈ 2.09 s. However, this is the period of oscillation for a small angle of displacement, and for larger angles, the period may be slightly different. Therefore, the answer is approximately 1 s.
57 . A spring-mass system is undergoing simple harmonic motion. If the amplitude of oscillation is doubled, what happens to the maximum speed of the mass?
a) It is halved
b) It is doubled
c) It remains the same
d) It is quadrupled
Answer: b) It is doubled
Explanation: The maximum speed of a mass in a spring-mass system undergoing simple harmonic motion is given by the formula vmax = Aω, where A is the amplitude of oscillation and ω is the angular frequency. If the amplitude is doubled, the maximum speed is also doubled, since vmax ∝ A.
58 . A system undergoing simple harmonic motion has a period of 4 s. What is its frequency?
a) 0.25 Hz
b) 0.5 Hz
c) 1 Hz
d) 2 Hz
Answer: b) 0.5 Hz
Explanation: The frequency of a system undergoing simple harmonic motion is given by the formula f = 1/T, where T is the period. Substituting the given value, we get f = 1/4 = 0.25 Hz.
59 . The displacement of a mass undergoing simple harmonic motion is given by x = 0.1 cos(2πt). What is its amplitude?
a) 0.1
b) 0.2
c) 0.3
d) 0.4
Answer: a) 0.1
Explanation: The displacement of a mass undergoing simple harmonic motion is given by the formula x = A cos(ωt), where A is the amplitude and ω is the angular frequency. Comparing this with the given equation, we see that A = 0.1.
60 . The time period of a mass-spring system is 2 s. If the mass is doubled and the spring constant is halved, what happens to the time period?
a) It is halved
b) It is doubled
c) It remains the same
d) It is quadrupled
Answer: c) It remains the same
Explanation: The time period of a mass-spring system is given by the formula T = 2π√(m/k), where m is the mass and k is the spring constant. If the mass is doubled and the spring constant is halved, the ratio of mass to spring constant remains the same, and therefore the time period remains the same.