Oscillations

JEE Main level, single correct, medium

148 of 194

Question 148 of 194

A simple pendulum of string length 30 cm performs 20 oscillations in 10 s. The length of the string required for the pendulum to perform 40 oscillations in the same time duration is ____\_\_\_\_ cm. [Assume that the mass of the pendulum remains same.]

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Answer: (A) 7.5

Period of simple pendulum: T=timenumber of oscillationsT = \frac{\text{time}}{\text{number of oscillations}}
Initial period: T1=1020=0.5T_1 = \frac{10}{20} = 0.5 s
Required period: T2=1040=0.25T_2 = \frac{10}{40} = 0.25 s
Using T=2πLgT = 2\pi\sqrt{\frac{L}{g}}, the ratio is: T2T1=L2L1\frac{T_2}{T_1} = \sqrt{\frac{L_2}{L_1}}
0.250.5=L230⇒0.5=L230\frac{0.25}{0.5} = \sqrt{\frac{L_2}{30}} \Rightarrow 0.5 = \sqrt{\frac{L_2}{30}}
0.25=L230⇒L2=7.50.25 = \frac{L_2}{30} \Rightarrow L_2 = 7.5 cm

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