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 2.70 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 (a) from x = 0.180 m to x = - 0.180 m and (b) from x = 0.090 m to x = - 0.090 m?

Solution 14E Step 1 of 4: In the given problem when the small block is attached to ideal spring.When the amplitude is A=0.09 m it takes t=2.7s to travel from x=0.09m to x=-0.09m. Now when the amplitude is doubled, that is A=0.18 m, we need to calculate the time taken by the block to travel (a) From x= 0.180m to x= -0.180 m (b) From x= 0.09m to x= -0.09m Step 2 of 4: Using the relation of angular frequency, = 2f = 2 T Where is angular frequency, f is frequency and T is time period From above equation, the frequency and time period does not depend on the amplitude. Hence time period will be same for both the case Time period is the time taken to complete one complete oscillation. In the given problem, it takes t=2.7s to complete half cycle, so for one complete cycle or oscillation T=2t T=2(2.7s) T=5.4 s Therefore, the time period of the given system is 5.4s.