A particle executing simple harmonic motion starts from mean position with amplitude ' A ' and periodic time ' T '. At what displacement is its speed one-fourth of the maximum spee

Question

A particle executing simple harmonic motion starts from mean position with amplitude ' A ' and periodic time ' T '. At what displacement is its speed one-fourth of the maximum speed?

TestHub · Accepted Answer

The velocity of a particle executing simple harmonic motion at a displacement $x$ from the mean position is given by $v = ω A^{2} - x^{2}$ .

The maximum velocity of the particle is $v_{ma x} = ω A$ .

Given that $v = \frac{v _{ma x}}{4}$ .

Substituting the expressions for $v$ and $v_{ma x}$ gives $ω A^{2} - x^{2} = \frac{ω A}{4}$ .

Squaring both sides yields $A^{2} - x^{2} = \frac{A ^{2}}{16}$ .

Rearranging the terms gives $x^{2} = A^{2} - \frac{A ^{2}}{16} = \frac{15 A ^{2}}{16}$ .

Taking the square root gives $x = \frac{A 15}{4}$ .

Answer: $\frac{A 15}{4}$

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