Question

A spring mass system oscillates with a frequency $v$. If it is taken in an elevator slowly accelerating upward, the frequency will:

• A. Increase
• B. Decrease
• C. Remain same
• D. Become zero

Answer 1

The correct option is A Increase

For spring mass $m$ system, the time period of the oscillation of mass $m$ is defined as
$T=2\pi \sqrt{m/k}$

When an elevator is accelerating upward (let say acc. a ) and eqm position of spring is $x_0$

Then $mg+ma=k{x}_0$

Now, when block is extended by a small distance $\Delta x$ Then restoring force $F^{\prime }=k(x_0+\Delta x)-mg-ma=k\Delta x$

By SHM $m{\omega }^2\Delta x=k\Delta x=\omega =\sqrt{\frac{k}{m}}⟹T=2\pi \sqrt{\frac{m}{k}}$ which is independent of acceleration .

So the time period will be same and hence frequency will remains same.