The maximum acceleration of particle executing simple harmonic oscillations is a and the maximum velocity is v. What is the frequency and amplitude of oscillation?

Solution

Let

straight omega

be the frequency and A the amplitude of oscillation of a particle executing simple harmonic oscillation.

Maximum acceleration is,

straight a subscript straight omicron equals straight omega squared straight A space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis

Maximum velocity is given by,

straight v subscript ring operator equals omega A space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis

Squaring (2) and dividing by(1), we get

straight v subscript 0 squared over straight a subscript 0 equals fraction numerator straight omega squared straight A squared over denominator straight omega squared straight A end fraction equals straight A therefore space space space straight A equals straight v subscript 0 squared over straight a subscript 0 comma space i straight s space the space amplitude space of space the space oscillation.

Also dividing (1) by (2), we have,

straight a subscript 0 over straight v subscript 0 equals fraction numerator straight omega squared straight A over denominator ωA end fraction equals straight omega Therefore comma space space space space space straight omega equals straight a subscript 0 over straight v subscript 0 comma space is space the space frequency space of space the space oscillation. space

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