- #1
Zolo
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There is a horizontal spring-mass system lie on top of a frictionless table.
let m=mass
k=spring constant
x=displacement from equilibrium position
A=amplitude
T=period
By using conventional way, time taken to complete a quarter of cycle=T/4=(∏/2)√(m/k)
consider another way of doing, F=-kx
a=-(k/m)x
then we can consider the average acceleration of first quarter of motion a=(1/2)(k/m)A
since s=ut+1/2at^2
then, A=(1/2)*(1/2)(k/m)A*t^2
finally, we get t=2√(m/k)
why is the ans from both way diffrent...Both ways seem to be equivalent...
let m=mass
k=spring constant
x=displacement from equilibrium position
A=amplitude
T=period
By using conventional way, time taken to complete a quarter of cycle=T/4=(∏/2)√(m/k)
consider another way of doing, F=-kx
a=-(k/m)x
then we can consider the average acceleration of first quarter of motion a=(1/2)(k/m)A
since s=ut+1/2at^2
then, A=(1/2)*(1/2)(k/m)A*t^2
finally, we get t=2√(m/k)
why is the ans from both way diffrent...Both ways seem to be equivalent...