Question

The time period of oscillation of a mass M, hanging from a spring of force constant K is T When additional mass m is attached to the spring, the period of oscillation becomes 5T/4. Then m/M IS

Answer 1

m/ M will become more as it is attached by extra mass

Answer 2

Given:

Mass hanging from the spring = M

Spring constant = K

Period of oscillation = T

New period of oscillation = $\frac{5T}{4}$

Mass which is added = m

To find:

Ratio  $\frac{m}{M}$

Formula Used:

Period of oscillation of spring mass system is given as follows:

$T=2\pi \sqrt{\frac{M}{K}$     ....(1)

Calculations:

When, mass m is added, new period becomes:

$\frac{5T}{4} = 2\pi \sqrt{\frac{m+M}{K}$     ....(2)

Divide (2) by (1), square both sides and solve:

$\frac{25}{16}=\frac{m+M}{M}\\\frac{25}{16}=\frac{m}{M}+1\\\frac{m}{M} = \frac{25}{16}-1\\\frac{m}{M}=\frac{9}{16}$

Thus, the ratio $\frac{m}{M}=\frac{9}{16}$

Learn more about: Spring-mass system

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