Question

A 1.0 kg mass at the end of a spring has an amplitude of 0.10 m and vibrates 2.0 times per second. What is its velocity when it is 0.10 m from equilibrium.

Answer 1

`v=0m/s`

Step-by-step explanation:

From the question we are told that

Mass of object
`M=10kg`

Amplitude
`A=0.10m`

SHM
`w=2.0 times per second.`

Distance from equilibrium
`0.10m`

Generally the equation for SHM displacement from equilibrium position is mathematically given by

`v=Aw\sqrt{1-(x^2)/(A^2)}`

For
`x=0.10`

Therefore

`v=Aw\sqrt{1-(0.10^2)/(A^2)}`

`v=0.10*2\sqrt{1-(0.10^2)/(0.10^2)}`

`v=0.10*2√(1-1)`

`v=0m/s`