Question

Figure 15-61shows that if we hang a block on the end of a spring with spring constant k, the spring is stretched by distance$\mathbf{h}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$. If we pull down on the block a short distance and then release it, it oscillates vertically with a certain frequency. What length must a simple pendulum have to swing with that frequency?

Short answer

The length of a simple pendulum which swings with the frequency of springis 2.0cm.

Step-by-step solution

The given data

Stretched distance h= 2.0 cm.

Understanding the concept of Hooke’s Law

Using the force of spring and force of gravity on a block, we can find whether this spring obeys Hook’s law or not. If this spring obeys Hook’s law, then using the formula of the period of a simple pendulum and the formula of frequency of oscillation, we can find the length of the simple pendulum which swings with the frequency of spring.

Formulae:

The force due to gravity,$F_g=mg$(i)

The force of a spring due to Hooke’s Law,$F_s=-kh$(ii)

The frequency of an oscillation,$f=\frac{1}{2\mathrm{\pi }}\sqrt{k/m}$(iii)

The time period of an oscillation, $T=\frac{1}{2\mathrm{\pi }}\sqrt{l/g}orT=\frac{1}{f}$ (iv)

Calculation of length of pendulum swinging with the frequency of pendulum

At equilibrium, force of spring + force of gravity on the block=net force is zeo

Therefore, using equations (i) and (ii), we get

$$\begin{gathered} mg-kh=0 \\ mg=kh \\ h=\frac{mg}{k} \end{gathered}$$

When spring stretched distance, new spring force is

$F_{net}=-kh-kx$

Therefore, net force is

$F_{net}=-kh-kx+mg$

Asrole="math" localid="1657281105057" $kh=mg,\mathrm{from}\mathrm{equation}\left(1\right),\mathrm{we}\mathrm{get}$

$$\begin{gathered} F_{net}\mathit{=}\mathit{-}kh\mathit{-}kx\mathit{+}kh \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}kx \end{gathered}$$

For simple pendulum, we have

$f\text{'}=\frac{1}{2\pi }\sqrt{\frac{g}{L}}$

Now, if the frequency is to be same with that of the frequency of equation (iii), then we can write,

role="math" localid="1657281377410" $$\begin{gathered} f=f\text{'} \\ \frac{1}{2\mathrm{\pi }}\sqrt{\frac{k}{m}}=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{g}{L}} \\ L=\frac{mg}{k} \\ =h(\mathrm{from}\mathrm{equation}(\mathrm{iv}\left)\right) \end{gathered}$$

Since,$h=2.0\mathrm{cm}$, therefore$L=2.0\mathrm{cm}$.

Therefore, the length of simple pendulum which swing with frequency of spring is 2 cm.