Question

If a mass M suspended by a spring executes SHM
with time period T, and the time period become
5T/3
if mass is increased by m. Then find ratio of
m/M

Answer 1

Ratio of $\frac{m}{M}$ will be $\frac{m}{M}=\frac{9}{16}$

Explanation:

We have given that mass of the spring is M

Time period = T

We know that time period of the spring is given by

$T=2\pi \sqrt{\frac{M}{k}}$----EQN 1

Now mass is increased by m

So new mass = M+m

New time period $T'=\frac{5T}{3}$----EQN 2

Dividing eqn 1 by eqn 2

$\frac{3}{5}=\sqrt{\frac{M}{M+m}}$

Now squaring both side

$\frac{9}{25}=\frac{M}{M+m}$

$9(M+m)=25M$

$\frac{m}{M}=\frac{9}{16}$

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Answer 2

Initally, the string is supporting mass M.

And , Force constant is K.

Time period of SHM can be given as :

T₀ = 2π√ (M/ k) ...(i)

Now, Mass changes to (M + m). then time period can be given as:

T =  2π√(M+m) / k

or, 5T / 3 = 2π√(M+m) / k  ...(ii)

Dividing eq(i) by eq(ii) we obtain:

5/3 = √(M+m)/M

Squaring the eqaution,

25/9 = (M+m) / M

or, m / M = 16 / 9

Therefore, the ratio of m/M is 16/ 9.

Hence, option b) 16/9 is correct.