HW-9 PHYS 101 FALL-17 Total points (25) Chapter 15 1. (points 2) MC If we double
ID: 1775436 • Letter: H
Question
HW-9 PHYS 101 FALL-17 Total points (25) Chapter 15 1. (points 2) MC If we double the frequency of a system undergoing simple harmonic motion, which of the following statements about that system are true? Please explain. (a) The period is doubled b The period is reduced to one-half of what it was Since, T=2@r/f c) The amplitude is doubled (d) The amplitude is reduced to one-half of what it was 2. (points 2) MC A simple harmonic oscillator consists of a mass oscillating on the end of a spring. Where is the acceleration of the system a maximum? Please explain. (a) At the amplitudes (b) At the equilibrium position (c) Where the kinetic energy is a maximum (d) Half way between the amplitudes (e) The acceleration is constant. 3. (points 4) A 1.50 kg mass on a spring has displacement as a function of time given by the equation x(t)-(740cm) cos [(416 s-1-2.42]. Find (a) the time for one complete vibration; (b) the force constant of the spring; (c) the maximum speed of the mass (d) the maximum force on the mass; (e) the position of the mass at t-1.00s (0 the force on the mass at that time. 4. (points 4) A 0.150-kg toy is undergoing SHM on the end of a horizontal spring with spring constant k-300 N/m. When the object is 0.0 120 m from its equilibrium position, it is observed to have a speed of 0.300 m/s. What are (a) the total energy of the object at any point of its motion; (b) the amplitude of the motion; (c) the maximum speed attained by the object during its motion? 5. (points 2) MC An oscillator creates periodic waves on a stretched string. If the amplitude of the oscillator doubles, what happens to the wavelength and wave speed? Please explain. (a) doubles, speed is halved (b) is unchanged, speed is halved (c) doubles, speed is unchanged (d) doubles, speed is doubled (e) wavelength and speed are unchanged Chapter 16Explanation / Answer
Mass m = 0.15 kg
Force constant k = 300 N / m
Displacement x = 0.012 m
Speed v = 0.3 m/ s
Angular frequency = [k / m]
= 44.72 rad / s
We know v = [A 2 - x 2 ]
[A 2 - x 2 ] =(v/) 2
= 4.5 x 10 -5
A 2 = x 2 + ( 4.5 x 10 -5 )
= 1.89 x 10 -4
A = 0.0137 m
(A) Total energy E = ( 1/ 2) m 2 A 2 = 0.0283 J
(B) Amplitude A = 0.0137 m
(C) maximum speed V = A = 0.613 m / s
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