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NEET PHYSICSOscillationsEasy

Question

Two simple harmonic motions of angular frequency 100 rad s1100\text{ rad s}^{-1} and 1000 rad s11000\text{ rad s}^{-1} have the same displacement amplitude. The ratio of their maximum acceleration will be:

A

1:101:10

B

1:1021:10^2

C

1:1031:10^3

D

1:1041:10^4

Step-by-Step Solution

  1. Identify the Formula: The maximum acceleration (amaxa_{max}) of a particle executing simple harmonic motion is given by the formula amax=ω2Aa_{max} = \omega^2 A, where ω\omega is the angular frequency and AA is the displacement amplitude .
  2. Analyze Given Data:
  • Angular frequency of first SHM, ω1=100 rad s1\omega_1 = 100\text{ rad s}^{-1}.
  • Angular frequency of second SHM, ω2=1000 rad s1\omega_2 = 1000\text{ rad s}^{-1}.
  • The amplitude (AA) is the same for both motions.
  1. Calculate Ratio: a1a2=ω12Aω22A=(ω1ω2)2\frac{a_{1}}{a_{2}} = \frac{\omega_1^2 A}{\omega_2^2 A} = \left(\frac{\omega_1}{\omega_2}\right)^2 a1a2=(1001000)2=(110)2=1100\frac{a_{1}}{a_{2}} = \left(\frac{100}{1000}\right)^2 = \left(\frac{1}{10}\right)^2 = \frac{1}{100}
  • The ratio is 1:1021:10^2.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Oscillations. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSOscillationssimpleharmonicmotionsangularfrequency

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