shown in the figure below. The block is initially at the equilibrium point at x = 0 m where the spring is at its natural length. Then the block is set into a simple harmonic oscillation with an initial velocity 2.5 m/s at x = 0 cm towards right. The horizontal surface is frictionless. a) What is the period of block’s oscillation?  b) Find the amplitude A of the oscillation, which is the farthest length spring is stretched to.  c) Please represent block’s motion with the displacement vs. time function x(t) and draw the motion graph x(t) for at least one periodic cycle. Note, please mark the amplitude and period in the motion graph. Assume the clock starts from when the block is just released.

Physics for Scientists and Engineers, Technology Update (No access codes included)
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Author:Raymond A. Serway, John W. Jewett
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Chapter1: Physics And Measurement
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A block of mass m = 2 kg is attached a spring of force constant k = 500 N/m as shown in the figure below.
The block is initially at the equilibrium point at x = 0 m where the spring is at its natural length. Then the
block is set into a simple harmonic oscillation with an initial velocity 2.5 m/s at x = 0 cm towards right. The
horizontal surface is frictionless.


a) What is the period of block’s oscillation? 


b) Find the amplitude A of the oscillation, which is the farthest length spring is stretched to. 


c) Please represent block’s motion with the displacement vs. time function x(t) and draw the motion graph
x(t) for at least one periodic cycle. Note, please mark the amplitude and period in the motion graph.
Assume the clock starts from when the block is just released. 


d) Please find out the block’s acceleration when it is at position x = 5cm. 


e) On the motion graph you draw for part c), please mark with diamonds ♦ where the kinetic energy of the
block is totally transferred to the spring elastic potential; mark square ■ where the elastic potential of the
spring is totally transferred to the block as kinetic energy. 

 

v = 2.5 m/s
wwwm-
m
x = 0
Transcribed Image Text:v = 2.5 m/s wwwm- m x = 0
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