A small block is attached to an ideal spring and is moving in SHM on a horizonta

Question

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 0.090 m, it takes the block 2.66 s to travel from x = 0.090 m to x = -0.090 m. If the amplitude is doubled, to 0.180 m, how long does it take the block to travel from x = 0.180 m to x = -0.180 m? Express your answer to two significant figures and include the appropriate units. If the amplitude is doubled, to 0.180 m, how long does it take the block to travel from x = 0.090 m to x = -0.090 m? Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

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

initially time period was 2*2.9=5.8 second.

i.e(2* going from -A to A) where A is amplitude.

hence afterwards also it will be same and time required from goingto

-A TO A will be same =2.9 SECONDS

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

=1.0833rad/second

time to reach 0.09

y=0.09

we get 0.09=0.18sin(1.08333t)

hence 1.0833t=pi/6.

hence t=0.4811seconds

but from going from 0.09 to -0.09

it will be 2*0.48111 seconds=0.96222seconds

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