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Simple Harmonic Motion question

2021 · 26 Feb · Shift 1 · Q53

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Simple Harmonic Motion
2021 · 26 Feb · Shift 1 · Q53

Simple Harmonic Motion question

2021 · 26 Feb · Shift 1 · Q53

JEE MainPhysicsSimple Harmonic MotionMCQ+4 / −1

If two similar springs each of spring constant K1 are joined in series, the new spring constant and time period would be changed by a factor :

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

View written solutionFree

Correct answer: C

Effective spring constant for two identical springs in series

If each spring has spring constant $K_1$ , then for two springs in series:

\frac{1}{K_{\text{eq}}} = \frac{1}{K_1} + \frac{1}{K_1} = \frac{2}{K_1}

So,

K_{\text{eq}} = \frac{K_1}{2}

Thus, the spring constant changes by a factor

\frac{K_{\text{eq}}}{K_1} = \frac{1}{2}

Effect on time period

For a mass-spring system,

T = 2\pi \sqrt{\frac{m}{K}}

Initially, with spring constant $K_1$ ,

T_1 = 2\pi \sqrt{\frac{m}{K_1}}

With the equivalent spring constant $K_{\text{eq}} = \frac{K_1}{2}$ ,

T_2 = 2\pi \sqrt{\frac{m}{K_1/2}}

T_2 = 2\pi \sqrt{\frac{2m}{K_1}} = \sqrt{2}\, T_1

So, the time period changes by a factor

\frac{T_2}{T_1} = \sqrt{2}

Match with options

The required factors are:

Spring constant factor: $\frac{1}{2}$
Time period factor: $\sqrt{2}$

This matches Option C.

Comparison with stored answer

Stored correct answer: C

Our derived answer: C

They agree.

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