An elastic string has a mass M suspended at its lower end- the upper being fixed to a support- When the mass is pulled down over a short distance and let go- then it executes SHM with Time period given by T -2 - l 1- g Where l 1 is the elongation of the string- Now the mass m is added to the mass M- then it is found that time period of oscillation T 2- such that T 1- T 2-5-4- Then the ratio 16 m - M is Given Y is the Young modulus of elasticity- L be the original length of string-

Let l_2 be the elongation in second case then nbsp; T_1 = 2 π√(l_1g) ....(1) T_2 = 2 π√(l_2g) ....(2) So that nbsp; T^2_1T^2_2 = l_1l_2 l_1l_2 = 1625 Now ...

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Standard XII

Physics

Elasticity

An elastic st...

Question

An elastic string has a mass $M$ suspended at its lower end, the upper being fixed to a support. When the mass is pulled down over a short distance and let go, then it executes SHM with Time period given by $T = 2 π \sqrt{\frac{l_{1}}{g}}$ Where $l_{1}$ is the elongation of the string. Now the mass $m$ is added to the mass $M$ , then it is found that time period of oscillation $T_{2},$ such that $\frac{T_{1}}{T_{2}} = \frac{5}{4}$ . Then the ratio $16 m : M$ is
Given $Y$ is the Young modulus of elasticity, $L$ be the original length of string.

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Solution

Let $l_{2}$ be the elongation in second case then
$T_{1} = 2 π \sqrt{\frac{l_{1}}{g}} . . . . (1)$
$T_{2} = 2 π \sqrt{\frac{l_{2}}{g}} . . . . (2)$
So that
$\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{l_{1}}{l_{2}}$

$\frac{l_{1}}{l_{2}} = \frac{16}{25}$
Now
for first case
$Y = \frac{M g L}{A l_{1}} . . . . (3)$
second case
$Y = \frac{(M + m) g L}{A l_{2}} . . . . (4)$
From 3 and 4
$\frac{l_{1}}{l_{2}} = \frac{M}{M + m}$

$\frac{M}{M + m} = \frac{16}{25}$
$\frac{M + m}{M} = \frac{25}{16}$
$\frac{M + m - M}{M} = \frac{25 - 16}{19}$
$\frac{m}{M} = \frac{9}{16}$

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Let l2 be the elongation in second case then T1=2π√l1g....(1) T2=2π√l2g....(2) So that T21T22=l1l2 l1l2=1625 Now for first case Y=MgLAl1....(3) second case Y=(M+m)gLAl2....(4) From 3 and 4 l1l2=MM+m MM+m=1625 M+mM=2516 M+m−MM=25−1619 mM=916