Question

At what distance from the mean position is the speed of a particle performing s.h.m. half its maximum speed. given path length of s.h.m. = 10 cm.​

Answer 1

Answer:

For an S.H.M. with angular frequency ω amplitude x and velocity v, at a distance x from the mean position, v^2 = ω^2(a^2 - x^2) . The maximum velocity is aω. Hence a^2ω^2/4 = ω^2(a^2 - x^2). Simplifying, 3a^2 = 4x^2 therefore

x = √3a/2. If in this case a = 10/2 = 5 cm, then x = 5√3/2 cm which is approximately 4.33 cm.

Explanation:

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Answer 2

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For an S.H.M. with angular frequency ω amplitude x and velocity v, at a distance x from the mean position, v^2 = ω^2(a^2 - x^2) . The maximum velocity is aω. Hence a^2ω^2/4 = ω^2(a^2 - x^2). Simplifying, 3a^2 = 4x^2 therefore

x = √3a/2. If in this case a = 10/2 = 5 cm, then x = 5√3/2 cm which is approximately 4.33 cm.