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Oscillations question

2008 · Q160
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Oscillations question

2008 · Q160

NEETPhysicsOscillationsMCQ+4 / −1
Two simple harmonic motions of angular frequency 100 and 1000 rad s−-−1 have the same displacement amplitude. The ratio of their maximum acceleration is
  1. A
    1:1031:{10^3}1:103
  2. B
    1:1041:{10^4}1:104
  3. C
    1:101:101:10
  4. D
    1:1021:{10^2}1:102
View written solutionFree

Correct answer: D

Maximum acceleration of a particle in the simple harmonic motion is directly proportional to the square of angular frequency i.e. a∝ω2a \propto {\omega ^2}a∝ω2

∴\therefore∴ a1a2=ω12ω22=(100)2(1000)2=1100{{{a_1}} \over {{a_2}}} = {{\omega _1^2} \over {\omega _2^2}} = {{{{\left( {100} \right)}^2}} \over {{{\left( {1000} \right)}^2}}} = {1 \over {100}}a2​a1​​=ω22​ω12​​=(1000)2(100)2​=1001​

⇒\Rightarrow⇒ a1 : a2 = 1 : 102.

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