A simple pendulum has time period ${T_1}$.The point of suspension is now moved upward according to the relation $y = k{t^2},\left( {k = 1{\rm{m}}{{\rm{s}}^{ - 2}}} \right)$ where y is the vertical displacement. The time period now becomes ${T_2}$ The ratio of $\frac{{T_1^2}}{{T_2^2}}$ is $\left( {{\rm{g}} = 10{\rm{m}}{{\rm{s}}^{ - 2}}} \right)$

JEE PHYSICS MCQ: OSCILLATIONS

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A simple pendulum has time period

{T_1}

.The point of suspension is now moved upward according to the relation

y = k{t^2},\left( {k = 1\mathrm{m}{\mathrm{s}^{ - 2}}} \right)

where y is the vertical displacement. The time period now becomes

{T_2}

The ratio of

\frac{{T_1^2}}{{T_2^2}}

\left( {\mathrm{g} = 10\mathrm{m}{\mathrm{s}^{ - 2}}} \right)

View answer and explanation

Correct option: A

6⁄5

Option A (6⁄5) is correct because the effective acceleration due to gravity changes when the suspension point moves upwards, affecting the time period.

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