Question

If the mass is displaced 0.270 m from equilibrium and released, what is its speed when it is 0.130 m from equilibrium?

Answer 1

Answer:

The speed is 1.52 m.

Explanation:

Given that,

Displacement =0.270 m

Distance = 0.130 m

Suppose a 0.321-kg mass is attached to a spring with a force constant of 13.3 N/m.

We need to calculate the angular velocity

Using formula of angular velocity

$\omega=\sqrt{\dfrac{k}{m}}$

Put the value into the formula

$\omega=\sqrt{\dfrac{13.3}{0.321}}$

$\omega=6.43\ rad/s$

We need to calculate the velocity

Using formula of velocity

$v=\omega\sqrt{A^2-x^2}$

Put the value into the formula

$v=6.43\times\sqrt{0.270^2-0.130^2}$

$v=1.52\ m/s$

Hence, The speed is 1.52 m.