Question

you are trying to obtain the maximum amplitude oscillation of a mass attached to a spring on a level surface (as in Fig 14-10). you pull the spring 0.1 m [right of the equailibrum position] and give it a has an initial velocity of 1 m/s [right] (i.e, move away from the equailibrum position). you measure the resulting amplitude of oscillation. you reset the mass at 0.1m [right] but this time give it a push so it has intial velocity of 1 m/s[LEFT] (i.e. towards equailibrum). Comaper the new amplitude of oscillation with the original one.

Answer 1

Here ,

let the amplitude of motion is A1

spring constant is k

Using energy conservation

0.5 * k * A^2 = 0.5 * m * v^2 + 0.5 * k * x^2

Now , for the second time

0.5 * k * A^2 = 0.5 * m * v^2 + 0.5 * k * x^2

hence , from the same equation ,

the amplitude will be same for both

the new amplitude of oscillation with the original one is 1