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A particle in S.H.M. crosses its equilib...

A particle in S.H.M. crosses its equilibrium position at time t = 2 sec. When the motion advances by one second more, its velocity is found to be `pi//3 m//s`. If the frequency of the motion is `1//6s^(10)`, find the amplitude of the motion in meter.

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To solve the problem, we need to find the amplitude of a particle in simple harmonic motion (SHM) given certain conditions. ### Step-by-Step Solution: 1. **Understand the Given Information:** - The particle crosses its equilibrium position at \( t = 2 \) seconds. - After 1 second (at \( t = 3 \) seconds), the velocity is \( v = \frac{\pi}{3} \, \text{m/s} \). - The frequency \( f = \frac{1}{6} \, \text{s}^{-1} \). ...
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Knowledge Check

  • A particle performing SHM starts equilibrium position and its time period is 16 seconds. After 2 seconds its velocity is pi m//s . Amplitude of oscillation is (cos 45^(@) = 1/(sqrt(2)))

    A
    `2sqrt(2) m`
    B
    `4sqrt(2) m`
    C
    `6sqrt(2) m`
    D
    `8sqrt(2) m`

  • The periodic time of a particle performing a linear S.H.M. is 12 sec. It starts from the mean position. After 2 seconds, its velocity is found to be pi cm//sec . What is the amplitude of S.H.M. ?

    A
    4cm
    B
    8cm
    C
    12 cm
    D
    16cm

  • A particle is performing a linear S.H.M. If at time t=0, its displacement is 1 cm, its velocity is pi cm//sec and its angular frequency is pi rad//s , then the amplitude of its motions is

    A
    1 cm
    B
    `sqrt(2) cm`
    C
    3 cm
    D
    `pi cm`