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 $1/Y$ is equal to

• A. $\begin{array}{c}[(\frac{T_M}{T})-1]\frac{Mg}{A}\end{array}$
• B. $\begin{array}{c}\\ [1-(\frac{T_M}{T}{)}^2]\frac{A}{Mg}\end{array}$
• C. $\begin{array}{c}\\ [1-(\frac{T}{T_M}{)}^2]\frac{A}{Mg}\end{array}$
• D. $\begin{array}{c}\\ [(\frac{T_M}{T}{)}^2-1]\frac{A}{Mg}\end{array}$

Answer 1

The correct option is D $\begin{array}{c}\\ [(\frac{T_M}{T}{)}^2-1]\frac{A}{Mg}\end{array}$

Initial time period $T=2\pi \sqrt{\frac{L}{g}}---\left(1\right)$
When the additional mass $M$ is added to the bob,
$T_M=2\pi \sqrt{\frac{l+\Delta l}{g}}$
$\Delta l$ is the extension in the wire due to mass $M$.
$\Delta l=\frac{Mgl}{AY}$
$\Rightarrow =2\pi \sqrt{\frac{l+\frac{Mgl}{AY}}{g}}---\left(2\right)$
$(2{)}^2÷(1{)}^2$
${\left(\frac{T_M}{T}\right)}^2=1+\frac{Mg}{AY}$
$\Rightarrow \frac{1}{Y}=\frac{A}{Mg}[{\left(\frac{T_M}{T}\right)}^2-1]$