For simple harmonic motion, if object 0.5kg is attached to a spring with force constant 8 n/m vibrates with SHM at amplitude 10cm with w(omega)=4;

calculate time for object to move from x = 8.00cm to x = 0 cm.

For this the explanation was explained using a x versus t cos graph in which the time for x = 8 and x = 0 was found, then the difference calculated.

However in practice, if its a spring in simple harmonic motion that oscillated with a 10cm amplitude, and if the spring wants to go from x = 8cm back down to x = 0 cm, doesn't that mean the spring has to complete the 10cm amplitude cycle first before going back down to 8cm? So the spring from x=8cm will have to go up to x=10cm and then back down to x=0 cm, and the time interval will be the time taken for x=8cm to x=10cm plus x=10cm to x=0cm?

Thankyou

## Waves and Oscillation 2

**Moderator:** msmod

### Re: Waves and Oscillation 2

I guess you assume it was on its way down from x = 8 to x = 0 rather than moving upwards from x = 8.