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NEET PHYSICSEasy

Two simple harmonic motions of angular frequencies 100100 and 1000 rad s−11000 \text{ rad s}^{-1} have the same displacement amplitude. The ratio of their maximum acceleration is:

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 Maximum Acceleration Formula: In Simple Harmonic Motion, the magnitude of maximum acceleration of a particle is given by amax=ω2Aa_{max} = \omega^2 A, where ω\omega is the angular frequency and AA is the amplitude .
  2. Set up Ratio: Let the two SHMs have angular frequencies ω1=100 rad s−1\omega_1 = 100 \text{ rad s}^{-1} and ω2=1000 rad s−1\omega_2 = 1000 \text{ rad s}^{-1}. They have the same amplitude, so A1=A2=AA_1 = A_2 = A.
  3. Calculate Ratio: Ratio=amax1amax2=ω12A1ω22A2\text{Ratio} = \frac{a_{max1}}{a_{max2}} = \frac{\omega_1^2 A_1}{\omega_2^2 A_2} Since A1=A2A_1 = A_2, this simplifies to: Ratio=(ω1ω2)2=(1001000)2\text{Ratio} = \left(\frac{\omega_1}{\omega_2}\right)^2 = \left(\frac{100}{1000}\right)^2 Ratio=(110)2=1100=1:102\text{Ratio} = \left(\frac{1}{10}\right)^2 = \frac{1}{100} = 1: 10^2
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