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.
$$\left[ {1 - {{\left( {\frac{{{T_M}}}{T}} \right)}^2}} \right]\frac{A}{{Mg}}$$
B.
$$\left[ {1 - {{\left( {\frac{T}{{{T_M}}}} \right)}^2}} \right]\frac{A}{{Mg}}$$
C.
$$\left[ {{{\left( {\frac{{{T_M}}}{T}} \right)}^2} - 1} \right]\frac{A}{{Mg}}$$
D.
$$\left[ {{{\left( {\frac{{{T_M}}}{T}} \right)}^2} - 1} \right]\frac{{Mg}}{A}$$
Answer :
$$\left[ {{{\left( {\frac{{{T_M}}}{T}} \right)}^2} - 1} \right]\frac{A}{{Mg}}$$
Solution :
As we know, time period, $$T = 2\pi \sqrt {\frac{\ell }{g}} $$
When additional mass $$M$$ is added then
$$\eqalign{
& {T_M} = 2\pi \sqrt {\frac{{\ell + \Delta \ell }}{g}} \cr
& {T_{\frac{M}{T}}} = \sqrt {\frac{{\ell + \Delta \ell }}{\ell }} \,\,{\text{or}}\,\,{\left( {\frac{{{T_M}}}{T}} \right)^2} = \frac{{\ell + \Delta \ell }}{\ell } \cr
& {\text{or,}}\,\,\,{\left( {\frac{{{T_M}}}{T}} \right)^2} = 1 + \frac{{Mg}}{{Ay}}\,\,\left[ {\because \Delta \ell = \frac{{Mg\ell }}{{Ay}}} \right] \cr
& \therefore \frac{1}{y} = \left[ {{{\left( {\frac{{{T_M}}}{T}} \right)}^2} - 1} \right]\frac{A}{{Mg}} \cr} $$