A small block is attached to an ideal spring and is moving i

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. When the amplitude of the motion is 0.090m , it takes the block 2.43s to travel from x= 0.090mto x= -0.090m . 1)If the amplitude is doubled, to 0.180m , how long does it take the block to travel from x= 0.180m to x= -0.180m ? 2)If the amplitude is doubled, to 0.180m , how long does it take the block to travel from x= 0.090m to x= -0.090m ?

Solution

1>frequency and time period doesn\'t depend upon amplitude
as T = 2(pi)*(m/k)^(0.5)

initially time period was 2*2.43=4.86 second.
i.e(2* going from -A to A) where A is amplitude.

hence afterwards also it will be same and time required from going to

-A TO A will be same =2.43 SECONDS

b>Y=Asin(wt)
A=0.18
w=2*pi*f=(2*pi)/T

=1.295rad/second

time to reach 0.09
y=0.09
we get 0.09=0.18sin(1.08333t)
hence 1.295t=pi/6.
hence t=0.404seconds

but from going from 0.09 to -0.09
it will be 2*0.404 seconds=0.808seconds