Question

When the spring, with the attached 275.0 g mass, is displaced from its new equilibrium position, it undergoes SHM. Calculate the period of oscillation, T , neglecting the mass of the spring itself.

Answer 1

The period of oscillation is 1.33 sec.

Step-by-step explanation:

Given that,

Mass = 275.0 g

Suppose value of spring constant is 6.2 N/m.

We need to calculate the angular frequency

Using formula of angular frequency

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

Where, m = mass

k = spring constant

Put the value into the formula

`\omega=\sqrt{(6.2)/(275.0*10^(-3))}`

`\omega=4.74\ rad/s`

We need to calculate the period of oscillation,

Using formula of time period

`T=(2\pi)/(\omega)`

Put the value into the formula

`T=(2\pi)/(4.74)`

`T=1.33\ sec`

Hence, The period of oscillation is 1.33 sec.