Answer :
Answer:
The new frequency will increase by a factor of [tex]\dfrac{1}{\sqrt2}[/tex].
Explanation:
The frequency of a simple pendulum is given by the formula as follows :
[tex]f=\dfrac{1}{2\pi }\sqrt{\dfrac{g}{l}}[/tex]
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
[tex]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[/tex]
So, the new frequency will increase by a factor of [tex]\dfrac{1}{\sqrt2}[/tex]. Hence, this is the required solution.
Answer:
Frequency will become [tex]\frac{1}{\sqrt{2}}[/tex] times
Explanation:
Let initially length of simple pendulum is L
Acceleration due to gravity is [tex]g=9.8m/sec^2[/tex]
Time period of the simple pendulum is to [tex]T=2\pi \sqrt{\frac{l}{g}}[/tex]
Now in second case length is doubled
So time period in second case [tex]T_{new}=2\pi \sqrt{\frac{2l}{g}}[/tex]
From the relation we can say that time period become [tex]\sqrt{2}[/tex] times
As frequency [tex]f=\frac{1}{T}[/tex]
So frequency will become [tex]\frac{1}{\sqrt{2}}[/tex] times