# JEE Main & Advanced Physics Simple Harmonic Motion Oscillation of Pendulum in Different Situations

Oscillation of Pendulum in Different Situations

Category : JEE Main & Advanced

(1) Oscillation in liquid : If bob a simple pendulum of density $\rho$ is made to oscillate in some fluid of density $\sigma$(where $\sigma <\rho$) then time period of simple pendulum gets increased.

As thrust will oppose its weight hence $m{{g}_{eff.}}=mg-$Thrust

or  ${{g}_{eff.}}=g-\frac{V\sigma g}{V\rho }$ i.e ${{g}_{eff.}}=g\,\left[ 1-\frac{\sigma }{\rho } \right]$

$\Rightarrow$ $\frac{g'}{g}=\frac{\rho -\sigma }{\rho }$

$\Rightarrow$ $\frac{T\,'}{T}=\sqrt{\frac{g}{g'}}=\sqrt{\frac{\rho }{\rho -\sigma }}>1$

(2) Oscillation under the influence of electric field : If a bob of mass m carries a positive charge q and pendulum is placed in a uniform electric field of strength E

(i) If electric field directed vertically upwards.

Effective acceleration

${{g}_{eff.}}=g-\frac{qE}{m}$

So  $T=2\pi \sqrt{\frac{l}{g-\frac{qE}{m}}}$

(ii) If electric field is vertically downward then

${{g}_{eff.}}=g+\frac{qE}{m}$ $T=2\pi \sqrt{\frac{l}{g+\frac{qE}{m}}}$

(3) Pendulum in a lift :  If the pendulum is suspended from the ceiling of the lift.

(i) If the lift is at rest or moving down ward /up ward with constant velocity.

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

and  $n=\frac{1}{2\pi }\sqrt{\frac{g}{l}}$

(ii) If the lift is moving up ward with constant acceleration a

$T=2\pi \sqrt{\frac{l}{g+a}}$ and $n=\frac{1}{2\pi }\sqrt{\frac{g+a}{l}}$

Time period decreases and frequency increases

(iii) If the lift is moving down ward with constant acceleration a

$T=2\pi \sqrt{\frac{l}{g-a}}$ and $n=\frac{1}{2\pi }\sqrt{\frac{g-a}{l}}$

Time period increase and frequency decreases

(iv) If the lift is moving down ward with acceleration  $a=g$

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

and $n=\frac{1}{2\pi }\sqrt{\frac{g-g}{l}}=0$

It means there will be no oscillation in a pendulum.

Similar is the case in a satellite and at the centre of earth where effective acceleration becomes zero and pendulum will stop.

(4) Pendulum in an accelerated vehicle : The time period of simple pendulum whose point of suspension moving horizontally with acceleration a

In this case effective acceleration ${{g}_{eff.}}=\sqrt{{{g}^{2}}+{{a}^{2}}}$

$T=2\pi \sqrt{\frac{l}{{{({{g}^{2}}+{{a}^{2}})}^{1/2}}}}$ and  $\theta ={{\tan }^{-1}}(a/g)\,\,$

If simple pendulum suspended in a car that is moving with constant speed v around a circle of radius r.

$T=2\pi \frac{\sqrt{l}}{\sqrt{{{g}^{2}}+{{\left( \frac{{{v}^{2}}}{r} \right)}^{2}}}}$

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