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 2.43 s to travel from x= 0.090 m to x= -0.090 m .

a) 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?

b) 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 ?

Answer 1

given
A = 0.090 m
T/2 = 2.43 s

T = 4.86 s

a) Time period does not depend on Amplitude.

so, even amplitude is doubled time taken for the block to travel from x= 0.180 m to x= -0.180 m is 2.43 s <<<<<<----Answer

b) angular frequency, w = 2*pi/T

= 2*pi/4.86

= 1.29 rad/s

let t is the time taken to move from x = 0.18 m to x = 0.090 m,

x = A*cos(w*t)

0.09 = 0.18*cos(1.29*t)

0.5 = cos(1.29*t)

t = cos^-1(0.5)/1.29

= 0.81178 s

time taken to move from x = -0.09m to x = -0.18 m is also 0.81178 s.

so, time taken to for the block to travel from x= 0.090 m to x= -0.090 m

delta_t = T/2 - 2*t

= 2.43 - 2*0.81178

= 0.806 s <<<<<<<---------------------Answer