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Grade: 11
        `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%
3 years ago

Answers : (1)

Rahul Verma
45 Points
							
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}     
 
T- 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%%
3 years ago
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