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Simple Harmonic Motion question

2017 · 8 Apr · Shift 1 · Q68

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JEE Main
Physics
Simple Harmonic Motion
2017 · 8 Apr · Shift 1 · Q68

Simple Harmonic Motion question

2017 · 8 Apr · Shift 1 · Q68

JEE MainPhysicsSimple Harmonic MotionMCQ+4 / −1

A 1 kg block attached to a spring vibrates with a frequency of 1 Hz on a frictionless horizontal table. Two springs identical to the original spring are attached in parallel to an 8 kg block placed on the same table. So, the frequency of vibration of the 8 kg block is :

A
${1 \over 4}Hz$
B
${1 \over {2\sqrt 2 }}Hz$
C
${1 \over 2}Hz$
D
$2Hz$

View written solutionFree

Correct answer: C

Original spring-block system

For a mass-spring system, the frequency is

f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}.

Given:

mass $m_1 = 1\,\text{kg}$
frequency $f_1 = 1\,\text{Hz}$

So,

1 = \frac{1}{2\pi}\sqrt{\frac{k}{1}}

which implies

\sqrt{k} = 2\pi.

We do not actually need the numerical value of $k$ ; only proportionality is enough.

New system with two identical springs in parallel

If two identical springs of spring constant $k$ are attached in parallel, the effective spring constant becomes

k_{\text{eq}} = k + k = 2k.

The new mass is

m_2 = 8\,\text{kg}.

Hence the new frequency is

f_2 = \frac{1}{2\pi}\sqrt{\frac{2k}{8}} = \frac{1}{2\pi}\sqrt{\frac{k}{4}}.

So,

f_2 = \frac{1}{2}\cdot \frac{1}{2\pi}\sqrt{k}.

But from the first case,

\frac{1}{2\pi}\sqrt{k} = 1.

Therefore,

f_2 = \frac{1}{2}\times 1 = \frac{1}{2}\,\text{Hz}.

Option check

A: $\frac{1}{4}\,\text{Hz}$ — incorrect
B: $\frac{1}{2\sqrt{2}}\,\text{Hz}$ — incorrect
C: $\frac{1}{2}\,\text{Hz}$ — correct
D: $2\,\text{Hz}$ — incorrect

Therefore, the correct answer is C.

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