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