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)

$\left[\left(\frac{T_{M}}{T}\right)^{2}-1\right] \frac{A}{Mg}$

Answer 2

$T =2 \pi \sqrt{\frac{\ell}{ g }} ; T _{ M }=2 \pi \sqrt{\frac{\ell'}{ g }}$
$\gamma=\frac{ Mg / A }{\Delta \ell / \ell}$
$\Rightarrow \frac{\ell'-\ell}{\ell}=\frac{ Mg }{\gamma A }=\frac{\ell'}{\ell}=1+\frac{ Mg }{\gamma A }$
Also:
$\frac{T_{M}}{T}=\sqrt{\frac{\ell'}{\ell}} \,\,\,\,\,\,\therefore T_{M}=T\left[1+\frac{M g}{\gamma A}\right]^{1 / 2}$
$\Rightarrow \frac{T_{M}^{2}}{T^{2}}=1+\frac{M g}{\gamma A}$
$\Rightarrow\left[\frac{T_{M}^{2}}{T^{2}}-1\right]=\frac{M g}{\gamma A}$
$\Rightarrow \frac{1}{\gamma}=\frac{A}{M g}\left[\left(\frac{T_{M}}{T}\right)^{2}-1\right]$