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PART - 2 CH - 13 Oscillations — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsPART - 2 CH - 13 Oscillations2 Marks
Question
Find the value of percentage change in the time period of the simple pendulum in the following situations :
(i) By increasing the length of pendulum by $5 \%$
(ii) On increasing the mass of the pendulum by $5\%$
(iii) By reducing the amplitude of pendulum by $5\%.$
✓

Answer

We know
$T=2 \pi \sqrt{\frac{l}{g}}\ldots\ldots (1)$
Here the length of pendulum is $l$ and the time period s T.
(i) On differentiation
$d T=\frac{2 \pi}{\sqrt{g}} \frac{1}{2 \sqrt{l}} d l\ldots\ldots (2)$
Equation (2) divided by (1)
$\begin{aligned}\frac{dT}{T} & =\frac{\frac{2 \pi}{\sqrt{g}} \frac{1}{2 \sqrt{l}} d l}{\frac{2 \pi}{\sqrt{g}} \sqrt{l}} \\\frac{dT}{T} & =\frac{d l}{2 l}\end{aligned}$
$\frac{dT}{T} \times 100 \%=\frac{1}{2}\left[\frac{d l}{l} \times 100 \%\right]$
Given : $\quad \frac{d l}{l} \times 100=5$
$\frac{dT}{T} \times 100=\frac{5}{2}$
Hence, percentage increase in time period is $=2.5 \%$.
(ii) The time period does not depend on the mass of the pendulum, hence it the mass in increased the time period will remain the same.
(iii) The period does not depend on the amplitude of the motion, hence if the amplitude is reduced the period will remain.
Hence, percentage change zero.

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