Question

If a simple pendulum oscillates with small amplitude and its length is doubled, what happens to the frequency of its motion? It doubles. It becomes 2 times as large. It becomes half as large. It becomes 1/ 2 times as large. It remains the same.

Answer 1

Answer:

Frequency will become $\frac{1}{\sqrt{2}}$ times

Explanation:

Let initially length of simple pendulum is L

Acceleration due to gravity is $g=9.8m/sec^2$

Time period of the simple pendulum is to $T=2\pi \sqrt{\frac{l}{g}}$

Now in second case length is doubled

So time period in second case $T_{new}=2\pi \sqrt{\frac{2l}{g}}$

From the relation we can say that time period become $\sqrt{2}$ times

As frequency $f=\frac{1}{T}$

So frequency will become $\frac{1}{\sqrt{2}}$ times

Answer 2

Answer:

The new frequency will increase by a factor of $\dfrac{1}{\sqrt2}$.

Explanation:

The frequency of a simple pendulum is given by the formula as follows :

$f=\dfrac{1}{2\pi }\sqrt{\dfrac{g}{l}}$

Here,

l is the length of the simple pendulum

g is acceleration due to gravity on which the pendulum is kept

The frequency of simple pendulum is independent of its amplitude. If a simple pendulum oscillates with small amplitude and its length is doubled, l' = 2l

$f'=\dfrac{1}{2\pi }\sqrt{\dfrac{g}{l'}}\\\\f'=\dfrac{1}{\sqrt2}\times \dfrac{1}{2\pi }\sqrt{\dfrac{g}{l}}\\\\f'=\dfrac{1}{\sqrt2}\times f$

So, the new frequency will increase by a factor of $\dfrac{1}{\sqrt2}$. Hence, this is the required solution.