Question

Time period of a simple pendulum is $T$. If its length increases by $2\%$, the new time period becomes

• A. 0.98 T
• B. 1.02 T
• C. 0.99 T
• D. 1.01 T

Answer 1

The Correct Option is D

$ T \propto l^{1 / 2} $ $ \Rightarrow \frac{\Delta T}{T} =\frac{1}{2}\left(\frac{\Delta l}{l}\right)$ $ \frac{\Delta T}{T} =\frac{1}{2}(2 \%)=1 \% $ $\Rightarrow \frac{T'-T}{T} =\frac{1}{100} $ $ \Rightarrow T'=T+0.01 \,T $ $ \Rightarrow T' =1.01 \, T$

Answer 2

Given that the mass of the bob is four times as great as the pendulum's length. to calculate the value of the time period.
A straightforward pendulum is made up of a string that is fastened to the bob of mass m at one end and to a pivot point at the other. The length of one complete back-and-forth swing is the basic pendulum's period. The vertical distance between the point of suspension and the center of the bob is what determines the length of the pendulum, which is nothing. It is symbolized by the letter "l".The basic pendulum's periodic time is given by,
\(T=2\pi\sqrt{\frac{l}{g}}\)
Where l is the length of the rope tied to the bob and g is the acceleration caused by gravity. The relationship between gravity and time is inverse. Therefore, the duration is shorter the larger the gravitational acceleration.

T₁=T,l₁=l,l₁=9l
\(\frac{T_1}{T_2}=\frac{\sqrt l_1}{\sqrt l_2}=\sqrt\frac{1}{9}=\frac{1}{3}\)
T₂=3T₁=3T
The time becomes three times if the pendulum's length is multiplied by nine.