13.Oscillations
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चंद्रमा के पृष्ठ पर गुरुत्वीय त्वरण $1.7\, m s ^{-2}$ है । यदि किसी सरल लोलक का पृथ्वी के पृष्ठ पर आवर्तकाल $3.5\, s$ है, तो उसका चंद्रमा के पृष्ठ पर आवर्तकाल कितना होगा ? (पृथ्वी के पृष्ठ पर $\left.g=9.8\, m\, s ^{-2}\right)$

A

$8.4$

B

$4.6$

C

$10.6$

D

$6.2$

Solution

Acceleration due to gravity on the surface of moon, $g^{\prime}=1.7\, m s ^{-2}$

Acceleration due to gravity on the surface of earth, $g=9.8\, m s ^{-2}$

Time period of a simple pendulum on earth, $T=3.5\, s$ $T=2 \pi \sqrt{\frac{l}{g}}$

Where, $l$ is the length of the pendulum

$\therefore l=\frac{T^{2}}{(2 \pi)^{2}} \times g$

$=\frac{(3.5)^{2}}{4 \times(3.14)^{2}} \times 9.8 \,m$

The length of the pendulum remains constant.

On moon's surface, time period, $T^{\prime}=2 \pi \sqrt{\frac{l}{g^{\prime}}}$ $=2 \pi \sqrt{\frac{\frac{(3.5)^{2}}{4 \times(3.14)^{2}} \times 9.8}{1.7}}=8.4\, s$

Hence, the time period of the simple pendulum on the surface of moon is $8.4\, s$

Standard 11
Physics

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