A 1.60-kg object oscillates from a vertically hanging light spring once every 0.
ID: 1472376 • Letter: A
Question
A 1.60-kg object oscillates from a vertically hanging light spring once every 0.53 s.
A) Write down the equation giving its position y (+ upward) as a function of time t (where t in seconds), assuming it started by being compressed 20 cm from the equilibrium position (where y=0), and released.
B)
How long will it take to get to the equilibrium position for the first time?
C)What will be its maximum magnitude of the speed?
D) What will be its maximum magnitude of the acceleration? where will the maximum magnitute of it be attained?
Explanation / Answer
a) T= 0.53s , f= 1/T= 1/0.53 = 1.89Hz , w= 2f = 2*3.14*1.89 = 11.9 rad/s , ym = 20cm = 0.2m , = 0
Equation of SHM,
y(t) = ymcos(wt + )
y(t) = 0.2cos(11.9t + 0 )
y(t) = 0.2cos(11.9t ) ----------(1)
b) y(t) = 0 m
From(1)
y(t) = 0.2cos(11.9t )
0= 0.2cos(11.9t )
0 = cos(11.9t )
cos^-1(0) = 11.9 t => t= 0.132s
c) Use equation,
vm = w*ym = 11.9*0.2 = 2.38 m/s
d) am = w^2*ym = 11.9*0.2^2 = 0.5 m/s^2
Maximum acceleration found at the extreme point i.e. where displacement is maximum and velocity is minimum.
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