Question 147: Oscillations

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 $${1 \over Y}$$ is equal to : 
($$g=$$ $$gravitational$$ $$acceleration$$)

Accepted Answer

As we know, time period, $$T = 2\pi \sqrt {{\ell  \over g}} $$
When a additional mass $$M$$ is added then
$${T_M} = 2\pi \sqrt {{{\ell  + \Delta \ell } \over g}} $$
$${{{T_M}} \over T} = \sqrt {{{\ell  + \Delta \ell } \over \ell }} $$
or, $$\,\,{\left( {{{{T_M}} \over T}} ight)^2} = {{\ell  + \Delta \ell } \over \ell }$$
or, $$\,\,{\left( {{{{T_M}} \over T}} ight)^2} = 1 + {{Mg} \over {Ay}}$$
$$\left[ \, ight.$$ as $$\left. {\Delta \ell  = {{Mg\ell } \over {Ay}}\,} ight]$$
$$	herefore$$ $${1 \over y} = \left[ {{{\left( {{{{T_M}} \over T}} ight)}^2} - 1} ight]{A \over {Mg}}$$

Answer

$$\left[ {1 - {{\left( {{{{T_M}} \over T}} \right)}^2}} \right]{A \over {Mg}}$$

Answer

$$\left[ {1 - {{\left( {{T \over {{T_M}}}} \right)}^2}} \right]{A \over {Mg}}$$

Answer

$$\left[ {{{\left( {{{{T_M}} \over T}} \right)}^2} - 1} \right]{A \over {Mg}}$$

Answer

$$\left[ {{{\left( {{{{T_M}} \over T}} \right)}^2} - 1} \right]{{Mg} \over A}$$

Oscillations

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

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