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Simple Harmonic Motion question

2006 · Shift 0 · Q112
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  5. /2006 · Shift 0 · Q112

Simple Harmonic Motion question

2006 · Shift 0 · Q112

JEE MainPhysicsSimple Harmonic MotionMCQ+4 / −1
The maximum velocity of a particle, executing simple harmonic motion with an amplitude 7mm,7mm,7mm, is 4.4m/s.4.4m/s.4.4m/s. The period of oscillation is
  1. A
    0.01s0.01s0.01s
  2. B
    10s10s10s
  3. C
    0.1s0.1s0.1s
  4. D
    100s100s100s
View written solutionFree

Correct answer: A

  1. Use the SHM relation for maximum velocity

For a particle in simple harmonic motion, vmax⁡=Aωv_{\max} = A\omegavmax​=Aω where:

  • AAA = amplitude
  • ω\omegaω = angular frequency

Also, ω=2πT\omega = \frac{2\pi}{T}ω=T2π​ So, vmax⁡=A⋅2πTv_{\max} = A\cdot \frac{2\pi}{T}vmax​=A⋅T2π​

  1. Substitute the given values

Amplitude: A=7 mm=7×10−3 mA = 7\text{ mm} = 7\times 10^{-3}\text{ m}A=7 mm=7×10−3 m

Maximum velocity: vmax⁡=4.4 m/sv_{\max} = 4.4\text{ m/s}vmax​=4.4 m/s

Thus, 4.4=7×10−3⋅2πT4.4 = 7\times 10^{-3}\cdot \frac{2\pi}{T}4.4=7×10−3⋅T2π​

  1. Solve for the period TTT

T=2πAvmax⁡T = \frac{2\pi A}{v_{\max}}T=vmax​2πA​

T=2π(7×10−3)4.4T = \frac{2\pi(7\times 10^{-3})}{4.4}T=4.42π(7×10−3)​

T=14π×10−34.4T = \frac{14\pi\times 10^{-3}}{4.4}T=4.414π×10−3​

Using π≈3.14\pi \approx 3.14π≈3.14, T=14×3.14×10−34.4T = \frac{14\times 3.14\times 10^{-3}}{4.4}T=4.414×3.14×10−3​

T=43.96×10−34.4T = \frac{43.96\times 10^{-3}}{4.4}T=4.443.96×10−3​

T≈9.99×10−3 sT \approx 9.99\times 10^{-3}\text{ s}T≈9.99×10−3 s

T≈0.01 sT \approx 0.01\text{ s}T≈0.01 s

  1. Match with the options

The correct option is: A: 0.01 s\boxed{\text{A: }0.01\text{ s}}A: 0.01 s​

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