Question

A $8kg$ mass is attached to a spring and allowed to hang in the Earths gravitational field. The spring stretches $2.4cm$ before it reaches its equilibrium position. If allowed to oscillate, what would be its frequency?

• A. $10Hz$
• B. $57.7Hz$
• C. $3.24Hz$
• D. $6.28Hz$

Answer 1

Answer: (C)

$3.24Hz$

For a spring-mass system, the angular frequency is given by $\omega =\sqrt{\frac{k}{m}}$

Thus, frequency is $f=\frac{\omega }{2\pi }=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$

Given that the spring stretches by $2.4\text{cm}$ when hanged.

As $kx=mg$ in equilibrium, we have $\frac{k}{m}=\frac{g}{x}=\frac{9.8}{2.4\times {10}^{-2}}=408.33{\text{s}}^{-2}$

Thus, frequency $f=\sqrt{408.33}/\left(2\pi \right)\approx =3.24\text{Hz}$