Question

The time period of a pendulum is given by $T=2\pi \sqrt{\frac{1}{g}}$ if the temperature rises by $△\theta$ and coefficient of linear expansion of the wire is $\alpha$, find out the change in time period is:

• A. $\alpha △\theta xT$
• B. $\frac{1}{2}\alpha △\theta T$
• C. $2\alpha △\theta xT$
• D. $\frac{1}{2}\alpha △\theta$

Answer 1

The correct option is B $\frac{1}{2}\alpha △\theta T$

Time period of pendulum (simple)$T=2\pi \sqrt{\frac{l}{g}}$

$l$=initial length of pendulum.

Now, coefficient of linear expansion of a wire $\alpha$ can be defined as,

$\alpha =\frac{increaseinlength}{temperaturechanged\times initiallength}$

$\Rightarrow$Increase in length$=\alpha \times$(Change in temperature) $\times$(Initial length)

$\Rightarrow$Increase in length$=\alpha \times △\theta \times l$

So, new length of wire$l^{\prime }=l+\alpha △\theta l=l(1+\alpha △\theta )$

So, now the time period is $T^{\prime }=2\pi \sqrt{\frac{l^{\prime }}{g}}$

$\sqrt{l^{\prime }}=\sqrt{l}(1+\alpha △\theta {)}^{\frac{1}{2}}$

As, $\alpha$ and $△\theta$ are very small, with respect to $1$

So, $(1+\alpha △\theta {)}^{\frac{1}{2}}\approx (1+\frac{1}{2}\alpha △\theta )$

We neglect the higher order term. So, $\sqrt{l}=\sqrt{l}(1+\frac{1}{2}\alpha △\theta )$

and, $T^{\prime }=2\pi \sqrt{\frac{l}{g}}(1+\frac{1}{2}\alpha △\theta )=T(1+\frac{1}{2}\alpha △\theta )$

$\Rightarrow T^{\prime }=T+\frac{1}{2}\alpha △\theta T$

$\Rightarrow T^{\prime }-T=Changeintimeperiod=\frac{1}{2}\alpha △\theta T$