By Valentina Tobos | Laurentiu Tobos

Updated on Sep 27, 2011

Review these concepts at: Simple Harmonic Motion Study Guide

### Practice Problems

- The elastic constant of a spring holding an object in equilibrium is 600 N/m. Find the extension of the spring with respect to its initial length if the object has a mass of 480 g.
- Draw a free-body diagram for an object hanging in equilibrium from a spring. Compare the magnitude of the forces you draw.

- If the period of an oscillatory motion is 2.4 seconds, what is the angular speed in rad/s and in degrees/s
- For the information given in practice problem 3, calculate the frequency and the angular displacement after
*T*/4,*T*/3,*T*/2, and*T* - Find the maximum acceleration and explain what the position is and what the speed of the object is when this happens.
- An object of mass 250 g oscillates by a spring and its motion is described by the following time dependence:
- For the information given in practice problem 6, determine the initial angular displacement at time
*t*= 0 s, the initial displacement, and speed.

*x*(*t*) = 1.2 m · sin (5 · *t* + 30°)

Determine the equations for the speed, acceleration, and force acting on the object as functions of time.

- At maximum displacement, an object in simple harmonic oscillation motion reaches the amplitude of 3.20 cm. The object is 1,250 g, and the mechanical energy of the object is measured to be 30 J. Find the maximum speed of the object.
- For the previous practice problem, find the angular frequency of the motion.
- Based on the information given in the graph in Figure 16.7, find the period, frequency, and angular frequency.
- Based on the information given in Figure 16.7, find the maximum potential energy and kinetic energy and the total energy of an object of mass 1 kg that describes an oscillatory motion as the one shown.
- Considering that between two consecutive positions the potential energy of an oscillator increases three times, what happens to the kinetic energy of the oscillator? Explain.
- Two completely identical oscillators are put in motion such that the amplitude of one is four times larger than the amplitude of the other, with all other parameters of motion being identical. Compare the total energies of the oscillators.

### Answers

- Δ
*x*= 12.5 mm - The two forces weight and elastic force are equal in magnitude and opposite in direction.
- 2.6 rad/s = 150 deg/s
- 0.42 Hz,
*π*/2 = 90°,2 ·*π*/3 = 120°,*π*= 180°, 2 ·*π*= 360° *a*(*t*) =*w*·.*A*· sin (*w*·*t*) and*a*_{max}=*w*•*A*when*t*=*T*/4 and*v*(*T*/4) = 0 m/s- Displacement is
*x*(*t*) = 1.2 m · sin (5 ·*t*+ 30°) and*v*(*t*) = 6 m/s · cos (5 ·*t*+ 30°) and acceleration is a(t) = –30 m/s^{2}• sin (5 ·*t*+ 30°) - At
*t*= 0*s*, angular displacement is θ(0*s*)· = 5 ·*t*+ 30° = 30°, initial displacement is*x*(0 s) = 1.2 m · sin (5 ·*t*+ 30°) = 0.6 m and initial speed is*v*(0*s*) = 6 m/s · cos (5 ·*t*+ 30°) = 5.2 m/s. *v*= 2 ·*E/m*·*A*^{2}= 15 m/s*v*=*w*^{2}·*A*,*w*· = (*v*/*A*)^{1/2}= 22 rad/s*T*= 8 s,*f*= 0.125 Hz,*w*· = 0.785 rad/s*PE*_{max}=*KE*_{max}=*E*= 31 J- The kinetic energy of the oscillator will decrease by the same amount the PE increases: Δ
*PE*=*Ef*–*Ei*= 3 ·*Ei*–*Ei*= 2 ·*Ei*, but*E*= constant so Δ*PE*= – Δ*KE*· Hence, Δ*KE*= – 2 ·*Ei*. - The oscillator moving with a larger amplitude will have a total mechanical energy four times larger.

From Physics Success in 20 Minutes A Day. Copyright © 2006 by LearningExpress, LLC. All Rights Reserved.

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