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

Answer:

$v=0m/s$

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

$v=Aw\sqrt{1-\frac{x^2}{A^2}}$

For $x=0.10$

Therefore

$v=Aw\sqrt{1-\frac{0.10^2}{A^2}}$

$v=0.10*2\sqrt{1-\frac{0.10^2}{0.10^2}}$

$v=0.10*2\sqrt{1-1}$

$v=0m/s$