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Simple Harmonic Motion question

2003 · Shift 0 · Q158

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Simple Harmonic Motion
2003 · Shift 0 · Q158

Simple Harmonic Motion question

2003 · Shift 0 · Q158

JEE MainPhysicsSimple Harmonic MotionMCQ+4 / −1

A mass

M

is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes

SHM

of time period

T.

If the mass is increased by

m.

the time period becomes

{{5T} \over 3}

. Then the ratio of

{{m} \over M}

is

A
${3 \over 5}$
B
${25 \over 9}$
C
${16 \over 9}$
D
${5 \over 3}$

View written solutionFree

Correct answer: C

For a mass–spring system, the time period is

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

where $M$ is the mass and $k$ is the spring constant.

When the mass is increased by $m$ , the new mass becomes

$(M+m)$

So the new time period is

$T' = 2\pi \sqrt{\frac{M+m}{k}}$

Given in the question:

$T' = \frac{5T}{3}$

Substitute the expressions for $T'$ and $T$ :

$2\pi \sqrt{\frac{M+m}{k}} = \frac{5}{3} \left(2\pi \sqrt{\frac{M}{k}}\right)$

Cancel $2\pi$ from both sides:

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

Square both sides:

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

Cancel $k$ :

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

Solve for $m$ :

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

Therefore,

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

Checking options:

A: $\frac{3}{5}$
B: $\frac{25}{9}$
C: $\frac{16}{9}$
D: $\frac{5}{3}$

So the correct option is C.

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