Question

URGENT PLEASE ANSWER ASAP!!!!Find the numerical part of your answer in standard notation and TWO significant digits. A 1.52 kg block is attached to a spring with a spring constant k = 89.0 N/m. It is set in motion by being displaced 9.22 cm downward. What is its frequency in Hz?

Answer 1

As we know that restoring force on the block when it is connected to a spring is given as

`F = -kx`

now we can say

`F = ma = - kx`

now we will have

`a = -(k)/(m) x`

now we can say that angular frequency of the motion is

`\omega^2 = (k)/(m)`

`\omega = \sqrt{(k)/(m)}`

`\omega = \sqrt(89)/(1.52)}`

`\omega = 7.65 rad/s`

now the frequency is given as

`f = (\omega)/(2\pi) = (7.65)/(2\pi)`

`f = 1.22 Hz`