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

Answer:

The period of oscillation is 1.33 sec.

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

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

Where, m = mass

k = spring constant

Put the value into the formula

$\omega=\sqrt{\dfrac{6.2}{275.0\times10^{-3}}}$

$\omega=4.74\ rad/s$

We need to calculate the period of oscillation,

Using formula of time period

$T=\dfrac{2\pi}{\omega}$

Put the value into the formula

$T=\dfrac{2\pi}{4.74}$

$T=1.33\ sec$

Hence, The period of oscillation is 1.33 sec.