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Simple Harmonic Motion question

2007 · Shift 0 · Q88
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Simple Harmonic Motion question

2007 · Shift 0 · Q88

JEE MainPhysicsSimple Harmonic MotionMCQ+4 / −1
Two springs, of force constant k1{k_1}k1​ and k2{k_2}k2​ are connected to a mass mmm as shown. The frequency of oscillation of the mass is f.f.f. If both k1{k_1}k1​ and k2{k_2}k2​ are made four times their original values, the frequency of oscillation becomes AIEEE 2007 Physics - Simple Harmonic Motion Question 146 English
  1. A
    2f2f2f
  2. B
    f/2f/2f/2
  3. C
    f/4f/4f/4
  4. D
    4f4f4f
View written solutionFree

Correct answer: A

  1. Effective spring constant

For a mass attached to springs, the frequency of small oscillation is

f=12πkeffm.f = \frac{1}{2\pi}\sqrt{\frac{k_{\text{eff}}}{m}}.f=2π1​mkeff​​​.

Regardless of the exact shown arrangement, the effective spring constant depends linearly on k1k_1k1​ and k2k_2k2​.

  • If springs are in parallel: keff=k1+k2k_{\text{eff}} = k_1 + k_2keff​=k1​+k2​
  • If springs are in series: keff=k1k2k1+k2k_{\text{eff}} = \frac{k_1k_2}{k_1+k_2}keff​=k1​+k2​k1​k2​​

In either case, if both k1k_1k1​ and k2k_2k2​ are multiplied by 444, then keffk_{\text{eff}}keff​ also gets multiplied by 444.

Let the new effective spring constant be keff′=4keffk'_{\text{eff}} = 4k_{\text{eff}}keff′​=4keff​.

  1. New frequency

The new frequency is

f′=12πkeff′m=12π4keffm=2⋅12πkeffm.f' = \frac{1}{2\pi}\sqrt{\frac{k'_{\text{eff}}}{m}} = \frac{1}{2\pi}\sqrt{\frac{4k_{\text{eff}}}{m}} = 2\cdot \frac{1}{2\pi}\sqrt{\frac{k_{\text{eff}}}{m}}.f′=2π1​mkeff′​​​=2π1​m4keff​​​=2⋅2π1​mkeff​​​.

Therefore,

f′=2f.f' = 2f.f′=2f.
  1. Option check
  • A: 2f2f2f ✅
  • B: f/2f/2f/2 ❌
  • C: f/4f/4f/4 ❌
  • D: 4f4f4f ❌

Hence, the correct answer is A.

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