Question 134: Oscillations

Question

A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of 'm' are attached at distance 'L/2' from its centre on both sides, it reduces the oscillation frequency by 20%. The value of radio m/M is close to :

Accepted Answer

Initially :
After putting 2 masses of each 'm' at a distance $${L \over 2}$$ from center :
We know,
Time period (T) = 2$$\pi $$ $$\sqrt {{{m I} \over C}} $$
$$ 	herefore $$  T $$ \propto $$ $$\sqrt {m I} $$
$$ 	herefore $$   Frequency (f) $$ \propto $$ $$\sqrt {{1 \over {m I}}} $$
$$ 	herefore $$   $${{{f_1}} \over {{f_2}}}$$ = $$\sqrt {{{{{m I}_2}} \over {{{m I}_1}}}} $$
Also given that,
After putting two masses 'm' at both end new frequency becomes 80% of initial frequency.
$$ 	herefore $$   f
2
= 0.8f
1
$$ 	herefore $$   $${{{f_1}} \over {0.8{f_1}}}$$ = $$\sqrt {{{{{m I}_2}} \over {{{m I}_1}}}} $$
$$ 	herefore $$   $${{{{{m I}_1}} \over {{{m I}_2}}}}$$ = 0.64
Initial moment of inertia of the system,
$${{{m I}_1}}$$ = $${{M{{\left( {2L} ight)}^2}} \over {12}}$$
Final moment of inertia of the system,
I
2
= $${{M{{\left( {2L} ight)}^2}} \over {12}}$$ + 2$$\left( {m{{\left( {{L \over 2}} ight)}^2}} ight)$$
$$ 	herefore $$   $${{M{{\left( {2L} ight)}^2}} \over {12}}$$ = 0.64 $$\left[ {{{M{L^2}} \over 3} + {{m{L^2}} \over 2}} ight]$$
$$ \Rightarrow $$   $${{M{L^2}} \over {3 	imes 0.64}}$$ = $${{M{L^2}} \over 3}$$ + $${{M{L^2}} \over 2}$$
$$ \Rightarrow $$   $${M \over {1.92}} - {M \over 3} = {m \over 2}$$
$$ \Rightarrow $$   $${{1.08M} \over {3 	imes 1.92}}$$ = $${m \over 2}$$
$$ \Rightarrow $$   $${m \over M}$$ = $${{1.08 	imes 2} \over {3 	imes 1.92}}$$ = 0.37

Answer

0.77

Answer

0.57

Answer

0.37

Answer

0.17

Oscillations

A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of 'm' are attached at distance 'L/2' from its centre on both sides, it reduces the oscillation frequency by 20%. The value of radio m/M is close to :

Choose the correct answer: