Question

A pendulum made of a uniform wire of cross sectional area $A$ has time period $T$. When an additional mass $M$ is added to its bob, the time period changes to $T_M$. If the Young's modulus of the material of the wire is $Y$ then $\frac{1}{Y}$ is equal to
( $g=$ gravitational acceleration)

• A. $\left[{\left(\frac{T_M}{T}\right)}^2-1\right]\frac{A}{Mg}$
• B. $\left[{\left(\frac{T_M}{T}\right)}^2-1\right]\frac{Mg}{A}$
• C. $\left[1-{\left(\frac{T_M}{T}\right)}^2\right]\frac{A}{Mg}$
• D. $\left[1-{\left(\frac{T}{T_M}\right)}^2\right]\frac{A}{Mg}$

Answer 1

Answer: (A)

$T=2\pi \sqrt{\frac{\ell }{g}};T_M=2\pi \sqrt{\frac{{\ell }^{\prime }}{g}}$
$\gamma =\frac{Mg/A}{\Delta \ell /\ell }$
$\Rightarrow \frac{{\ell }^{\prime }-\ell }{\ell }=\frac{Mg}{\gamma A}=\frac{{\ell }^{\prime }}{\ell }=1+\frac{Mg}{\gamma A}$
Also:
$\frac{T_M}{T}=\sqrt{\frac{{\ell }^{\prime }}{\ell }}\therefore T_M=T{\left[1+\frac{Mg}{\gamma A}\right]}^{1/2}$
$\Rightarrow \frac{T_M^2}{T^2}=1+\frac{Mg}{\gamma A}$
$\Rightarrow \left[\frac{T_M^2}{T^2}-1\right]=\frac{Mg}{\gamma A}$
$\Rightarrow \frac{1}{\gamma }=\frac{A}{Mg}\left[{\left(\frac{T_M}{T}\right)}^2-1\right]$