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A pendulum is executing simple harmonic motion and its maximum kinetic energy is K₁. If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is K₂. Then :

Option 1) K₂=K₁

Option 2) K₂=2K₁

Option 3) K₂=K₁/2

Option 4) K₂=K₁/4

Answers (1)

best_answer

Time period of oscillation of simple pendulum -

$T=2\pi \sqrt{\frac{l}{g}}$

- wherein

l = length of pendulum

g = acceleration due to gravity.

Kinetic energy in S.H.M. -

$K.E.= \frac{1}{2}mu^{2}\\ \: \: \: = \frac{1}{2}m\left ( A^{2} -x^{2}\right )\omega ^{2}$

- wherein

$K.E.= \frac{1}{2}K\left ( A^{2}-x^{2} \right ) \\K= m\omega ^{2}$

K.E will be maximum at point B.

& $KE=\frac{1}{2}mw^{2}A^{2}$

here $w=\sqrt{\frac{g}{L}}$

& $A=L\theta$

Now we double length keeping A same

So ${L}'=2L$

So ${A}'=A$

${A}'={L}'{\theta }'$

${\theta }'=\frac{\theta }{2}$

$KE=\frac{1}{2}mw^{2}A^{2}$

$KE=\frac{1}{2}m\times \frac{g}{L}\times A^{2}$

$\frac{K_{1}}{K_{2}}=\frac{L_{2}}{L_{1}}=\frac{2L_{1}}{L_{1}}$

$K_{2}=\frac{K_1}{2}$

Option 1)

K₂ = K₁

Option 2)

K₂ = 2K₁

Option 3)

$K_{2} = \frac{K_{1}}{2}$

Option 4)

$K_{2} = \frac{K_{1}}{4}$

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