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Physics & Astronomy

Quiz 13 :

Vibrations and Waves

Quiz 13 :

Vibrations and Waves

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A particle in simple harmonic motion has a frequency of 40 Hz. What is the period of this oscillation
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Time period is defined as the time taken for one complete oscillation. It is equal to the inverse of the frequency of the oscillation.
The relationship between time period
img and frequency of oscillation
img is,
img Substitute
img for
img in equation
img .
img Therefore, the period of oscillation is
img .

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If the amplitude of a particle in SHM is doubled, how are (a) the total energy and (b) the maximum speed affected
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(a)
The expression for the total energy of the particle executing Simple harmonic motion is,
img Here, m is the mass of the particle,
img is the angular frequency of oscillation, and
img is the amplitude of the particle.
From the above expression, keeping the other parameters constant, it is clear that the total energy of the particle in the simple harmonic motion is directly proportional to the square of the amplitude of the particle.
img Therefore, if the amplitude of a particle in the simple harmonic motion is doubled, then the total energy of the particle will be four times its initial energy.
(b)
The expression for the maximum speed of the particle in simple harmonic motion is,
img Since, the maximum speed of the particle is directly proportional to the amplitude of the particle. Therefore, if the amplitude of the particle is doubled then the speed of the particle will also be doubled.

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When a particle in a horizontal SHM is at the equilibrium position, the kinetic energy of the system is (a) zero, (b) at a maximum, (c) half the maximum value, (d) none of the preceding.
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A particle exhibits horizontal simple harmonic motion with amplitude
img ; the total energy of the particle is equal to kinetic and potential energies of the particle.
img Here,
img is force constant or spring constant,
img and
img are the mass and speed of the particle and
img is the position of the particle.
(a)
When the oscillating particle passes through its equilibrium position, then in that case the value of
img is zero and thus its potential energy is zero. At that instant, the total energy is equal to the kinetic energy and the object is traveling at its maximum speed
img .
Hence, total energy of an object in simple harmonic motion of a spring can be calculated as follows:
img In this case, at equilibrium position only kinetic energy is available as whole potential energy is converted into kinetic energy. Thus, at equilibrium position in a horizontal SHM, the kinetic energy of the system is at maximum.
In option (a), at equilibrium position, the kinetic energy of the system is mentioned as zero but as explained above, the kinetic energy of the system is calculated to be maximum at equilibrium position.
Hence, option (a) is incorrect.
(b)
In option (b), at equilibrium position, the kinetic energy of the system is mentioned to be at a maximum and in the section (a) above also, it is calculated that the kinetic energy of the system is maximum at equilibrium position.
Hence, option (b) is correct.
(c)
In option (c), at equilibrium position, the kinetic energy of the system is mentioned as half the maximum value but in the section (a), it is calculated that the kinetic energy of the system at equilibrium position is maximum and its value is
img .
Hence, option (c) is incorrect.
(d)
In option (d), it is mentioned in the problem that none of the above parts (a), (b) and (c) are correct but as explained above, the section (b) is correct.
Hence, option (d) is incorrect.

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If the frequency of a system in SHM is doubled, the period of the system is (a) doubled, (b) halved, (c) four times as large, (d) one-quarter as large.
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If it takes a particle in SHM 0.50 s to travel from the equilibrium position to the maximum displacement (amplitude), what is the period of oscillation
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For the SHM equation y = A sin[(200 rad/s) t ], the frequency of oscillation, f , is (a) 50 Hz, (b) 100 Hz, (c) 200 Hz, (d) 200 Hz.
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How does the speed of a mass in SHM change as the mass leaves its equilibrium position Explain.
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A particle oscillates in SHM with an amplitude A. What is the total distance (in terms of A ) the particle travels in three periods
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A tennis player uses a racket to bounce a ball up and down with a constant period. Is this a simple harmonic motion Explain.
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An object of mass 1.0 kg is attached to a spring with spring constant 15 N/m. If the object has a maximum speed of 0.50 m/s, what is the amplitude of oscillation
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For a particle in SHM, the force on it ( F ) and its displacement from its equilibrium position ( x ) are (a) in the same direction, (b) opposite in direction, (c) perpendicular to each other, (d) none of the preceding.
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If a mass-spring system were taken to the Moon, would the period of the system change How about the period of a pendulum taken to the Moon Explain.
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If the length of a pendulum is doubled, what is the ratio of the new period to the old one
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The equation of motion for a particle in SHM (a) is a sine or cosine function, (b) is a tangent or cotangent function, (c) could be any mathematical function, (d) gives the velocity of the particle as a function of time.
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If you want to increase the frequency of vibration of a mass-spring system, would you increase or decrease the mass Explain.
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A 0.75-kg object oscillating on a spring completes a cycle every 0.50 s. What is the frequency of this oscillation
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The frequency of a simple harmonic oscillator is doubled from 0.25 Hz to 0.50 Hz. What is the change in its period
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The maximum kinetic energy of a mass-spring system in SHM is equal to (a) A , (b) A 2 , (c) kA , (d) kA 2 /2.
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A mass-spring system in SHM has an amplitude A and period T. How long does the mass take to travel a distance A How about 2 A
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An object of mass 0.50 kg is attached to a spring with spring constant 10 N/m. If the object is pulled down 0.050 m from the equilibrium position and released, what is its maximum speed
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