91影视

Frequency is defined as the number of oscillations done by an object per second.

The frequency of this simple pendulum of length is given by

$\mathrm{f}=\frac{1}{2\mathrm{蟺}}\sqrt{\frac{\mathrm{g}}{\mathrm{L}}}$

Here, F is the frequency, g Is the acceleration due to gravity, and L is the length.

Consider the given data as below.

The weight of the apple is $\mathrm{W}=1\mathrm{N}$.

Spring force constant is $\mathrm{k}=1.5\mathrm{N}/\mathrm{m}$.

Acceleration due to gravity,$\mathrm{g}=9.8\mathrm{m}{\mathrm{s}}^{-2}$

Since the frequency of the mass attached with a spring having a spring constant k in SHM is given by

${\mathrm{f}}_{\mathrm{B}}=\frac{1}{2\mathrm{蟿蟿}}\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$ 鈥�.. (1)

Here,${\mathrm{f}}_{\mathrm{B}}$ is the bounce frequency.

Now stop the bouncing and let the apple swing from side to side through a small angle, in this case, the frequency of this simple pendulum of length L is given by

${\mathrm{f}}_{\mathrm{p}}=\frac{1}{2\mathrm{蟺}}\sqrt{\frac{\mathrm{g}}{\mathrm{L}}}$ 鈥�.. (2)

Here, ${\mathrm{f}}_{\mathrm{P}}$is the pendulum frequency and g is gravitational acceleration.

Since it is given that in the question that, frequency of simple pendulum is half the bounce frequency,

Hence from equations (1) and (2),

$$\begin{gathered} {\mathrm{f}}_{\mathrm{P}}=\frac{1}{2}{\mathrm{f}}_{\mathrm{B}} \\ \frac{1}{2\mathrm{蟿蟿}}\sqrt{\frac{\mathrm{g}}{\mathrm{L}}}=\frac{1}{2}\frac{1}{2\mathrm{蟿蟿}}\sqrt{\frac{\mathrm{k}}{\mathrm{m}}} \end{gathered}$$

$\mathrm{L}=\frac{4\mathrm{mg}}{\mathrm{k}}$ 鈥�.. (3)

Since the weight is,

$$\begin{gathered} \mathrm{W}=\mathrm{mg} \\ \mathrm{m}=\frac{\mathrm{W}}{\mathrm{g}} \end{gathered}$$

Substitute this into equation (3).

$$\begin{gathered} L=\frac{4\left(\frac{W}{g}\right)g}{k} \\ =\frac{4\mathrm{W}}{\mathrm{k}} \\ =\frac{4脳1}{1.5} \\ =2.67\mathrm{m} \end{gathered}$$

The extended length L is equal to the natural length${\mathrm{L}}_0$ plus x is the length of extension due to the weight of the apple.

Here, the length of extension due to the weight of the apple is,

$$\begin{gathered} \mathrm{x}=\frac{\mathrm{W}}{\mathrm{k}} \\ =\frac{1}{1.5} \\ =0.667\mathrm{m} \end{gathered}$$

Hence, the extended length will be

$$\begin{gathered} \mathrm{L}={\mathrm{L}}_0+\mathrm{x} \\ 2.67={\mathrm{L}}_0+0.67 \\ {\mathrm{L}}_0=2\mathrm{m} \end{gathered}$$

Hence, the unstretched length of the spring is 2 m.