Question

A small mass m is hung at the end of a light string of length 1.5 m and allowed to swing over a small amplitude as a simple pendulum. It is observed that the amplitude of swing is reduced to half its initial value in 30 complete swings. It may be assumed that the reduction of amplitude is due entirely to resistance of the air and this in turn may be assumed to be proportional to the velocity of the mass. Deduce in how many swings the amplitude would be reduced to half its initial value if the length of the pendulum were changed to 2 m. First show that the equation of motion for this system is: 0 +

Answer 1

The situation is as shown in the figure above.

The total force on the pendulum

$F=-mgsin(\theta )-Cv$

As

$\theta\approx 0$ and $v=L \dot\theta$

$F=-mg\theta -CL\dot{\theta}$

$F=mL\ddot\theta$

$mL\ddot\theta= -mg\theta-CL\dot{\theta}$

$\ddot\theta= -\frac{g}{L}\theta-\frac{C}{m}\dot{\theta}$

$\ddot\theta+\frac{C}{m}\dot{\theta} +\frac{g}{L}\theta=0$

This is the require equation of motion.It is second order total differential equation with constat coefficients.

d= Aexpl{l-C/2m)+ (C/2m)2 - g/l)t)+Berp(((-C/2m)- (C/2m)2 – 9/11)

the solution remains periodic for sufficient long only if

$(\frac{C}{2m})^{2}< < \frac{g}{L}$

which defnitely should be the case here.

Under this assumption the solution can be written as

$\theta=D_{0}exp((-C/2m)t)Cos(\sqrt{g/L}t+\phi )$

Hence The amplitude

$Amplitude ,A=D{_{0}} exp(\frac{-C}{2m})t$

decrease exponentially

with g=9.8 ms-2 and L=1.5 m

$period,T_{1.5}=2\pi\sqrt{} {L/g}=2\pi\sqrt{1.5/9.8}\approx 2.46 sec$

time taken to comple 30 swings

T30=73.8 sec

Amplitude after 30 swings =

.$A_{30}=D{_{0}} exp(\frac{-C}{2m})73.8$

$A_{30}=\frac{D_{0}}{2}$

$1/2=exp(\frac{-36.9C}{m})$

$ln(1/2)=\frac{-36.9C}{m}$

$\frac{C}{m}=.693/36.9=0.0188$

Let n  be the no of swings the amplitude will reduce by half if L=2.0m

$period,T_{2m}=2\pi\sqrt{} {L/g}=2\pi\sqrt{2/9.8}\approx 2.838 sec$

Time for n swings

Tn=2.838 n

soliving

$Amplitude ,A=D{_{0}} exp(\frac{-C}{2m})t$

with t=2,838n

C/m =0.0188

A=D0/2

we have

$\frac{1}{2}= exp(\frac{-0.0188}{2}2.828n)$

$n\approx 8$

Thus the no of swings is 8