3. An object is oscillating in SHM with amplitude of 2.5 cm and a period of 0.02
ID: 2306557 • Letter: 3
Question
Explanation / Answer
3.
a)
For SHM
Y=ASin(Wt)
angular frequency
W=2pi/T =2pi/0.02 =100pi rad/s
Given Y=0.3 cm
=>0.3=2.5Sin(100pi*t)
100pi*t =Sin-1(0.3/2.5)
t=3.83*10-4s
Velocity
V=dY/dt =AWCos(Wt)
V=0.025*100pi*Cos(100pi*3.84*10-4)
V=7.8 m/s
acceleration
a=dV/dt =-AW2Sin(Wt)=-W2Y =-(100pi)2*(0.3*10-2)
a=-296.1 m/s2
b)
Spring Constant
K=W2m =(100pi)2*0.025=2467.4 N/m
Restoring force
F=KY=2467.4*0.3*10-2
F=7.4 N
c)
Cyclic frequency of the object
W=100pi rad/s or 314.16 rad/s
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