Question

A mass hanging from a spring oscillates with a period of 0.35 s. Suppose the mass and spring are swung in a horizontal circle, with the free end of the spring at the pivot. What rotation frequency, in rpm, will cause the spring’s length to stretch by 15%?

Answer 1

Answer:

66 rpm

Explanation:

The period of oscillation is given by

$T=2\pi \sqrt{\frac {m}{k}}$

$\frac {k}{m}=\frac {4\pi^{2}}{T^{2}}$ where  T is time period of oscillation which is given as 0.35 s, k s spring constant and m is the mass of the object attached to the spring.

Also, net force is given by

Net force=$m\omega^{2} L$

$\omega=\sqrt{\frac {k\triangle L}{mL}}$ where $\triangle L$ is the elongation, L is original length, $\omega$ is the angular velocity

Substituting the equation of $\frac {k}{m}$ into the above we obtain

$\omega=\sqrt {\frac {4\pi^{2}\triangle L}{T^{2} L}}$

$\omega=\sqrt {4\pi^{2}\times 0.15L}{0.35^{2}\times L}}=6.952763\approx 6.95 rad/s$

$6.95\times\frac {60 s}{2\pi rad}\approx 66 rpm$