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`A simple pendulum is executing simple harmonic motion with a time period T; If the length of the pendulum. Is increased by 21%, the Increase in the time period of the pendulum of Increased length is `-(1)1o% (2)21% (3) 30% (4) 50%

Nitin Kumar , 9 Years ago

Grade 11

2 Answers

Rahul Verma

By formula $T_{1}=2\pi \sqrt{\frac{l}{g}}$ \

now length is increased by 21% therefore new length = $l+l\frac{21}{100}= \frac{121l}{100}$ put in eq we get

$T_{2}^{}= 2\pi \sqrt{\frac{\frac{121l}{100}}{g}}$ $T_{2}= 2\pi \sqrt{\frac{\frac{121l}{100}}{g}}=2\pi\sqrt{} \frac{121l}{100g}=2\pi \sqrt{\frac{l}{g}}\frac{11}{10}=T_{1}\frac{11}{10}$

T₂- T₁ = $\frac{T_{1}}{10}$

so percentage change = $\frac{change in time period }{original time period}\times 100$ $= \frac{\frac{T_{1}}{10}}{T_{1}}\times 100$ $=\frac{T_{1}}{10T_{1}}\times 100$ $= 10%$ %

Approved

Last Activity: 9 Years ago

Yash Chourasiya

Dear Student

Now, time period of simple pendulum,
l = length of simple pendulum
g = acceleration due to gravity
T = 2π√l2/g
If T₂ = Final time period & T₁ = Initial time period

Therefore percentage increase in time period = (T₂- T₁/T)x100%
643-143_1 (1).PNG

643-143_1 (1).PNG

I hope this answer will help you.
Thanks & Regards
Yash Chourasiya

Last Activity: 5 Years ago

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