(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.As the altitude increases,the acceleration due to gravity $g$ decre

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(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.As the altitude increases,the acceleration due to gravity $g$ decreases $(g' = g(1 - \frac{2h}{R}))$.Since $T \propto \sqrt{\frac{1}{g}}$,a decrease in $g$ leads to an increase in the time period $T$,causing the clock to run slow.To maintain the same time period $T$,the length $l$ of the pendulum must be reduced proportionally to the decrease in $g$.

Class 11 Physics Oscillations SHM of Simple Pendulum

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$A$ pendulum clock is set to give correct time at sea level. This clock is moved to a hill station at an altitude of $2500 \ m$ above sea level. In order to keep correct time at the hill station,the length of the pendulum:

A
Has to be reduced
B
Has to be increased
C
Needs no adjustment
D
Needs no adjustment but its mass has to be increased

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SHM of Simple Pendulum

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$A$ pendulum clock is set to give correct time at sea level. This clock is moved to a hill station at an altitude of $2500 \ m$ above sea level. In order to keep correct time at the hill station,the length of the pendulum:

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$A$ small sphere oscillates simple harmonically in a watch glass whose radius of curvature is $1.6 \text{ m}$. The period of oscillation of the sphere is (Acceleration due to gravity $g = 10 \text{ m/s}^2$)

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