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

(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.

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