Question

The figure shows that if we hang a block at the end of a spring with spring constant $k$, the spring is stretched by distance $h=2.0\text{cm}$. If we pull down the block a short distance and then release it, it oscillates vertically with a certain frequency. What length must a simple pendulum have, so that it swings with the same frequency?

• A. $1.0\text{cm}$
• B. $2.0\text{cm}$
• C. $3.0\text{cm}$
• D. $4.0\text{cm}$

Answer 1

The correct option is B $2.0\text{cm}$

Using Hooke's law, we have,
$mg=kh$ .................$\left(1\right)$
at equilibrium condition.

The frequency of oscillation for the mass-spring system is
$f=\frac{1}{T}=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$

Similarly, the frequency of oscillation for a simple pendulum is
$f^{{}^{\prime }}=\frac{1}{T^{{}^{\prime }}}=\frac{1}{2\pi }\sqrt{\frac{g}{L}}$

If $f=f^{{}^{\prime }}$, then,
$\frac{1}{2\pi }\sqrt{\frac{k}{m}}=\frac{1}{2\pi }\sqrt{\frac{g}{L}}$, which gives
$L=\frac{mg}{k}=\frac{kh}{k}$ $[$from $\left(1\right)]$
$\Rightarrow L=h=2.0\text{cm}$

Hence, option $\left(b\right)$ is correct.