Question

A body of mass 5 gram is executing S.H.M. about a fixed point O. With an amplitude of 10 cm, its maximum velocity is 100 cm/s. Its velocity will be 50 cm s–1 at a distance (in cm)(a) 5(b) 5√2(c) 5√3(d) 10√2

Answer 1

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Answer 2

Answer:

C) 5√3cm

Explanation:

Let the spring constant of spring = K

Velocity at mean position vm = 100cm

Amplitude A = 10cm

Applying conservation of energy at mean position and at the extreme position = K.E+P.E =K.Ea+P.Ea

= 1/2mv²m+0 = 0+1/2KA²

= 1/2×5×(100)²+0 = 0+1/2K(10)²

= 500cm

Velocity of body is 50cm at point b.

Thus,

= K.E+P.E =K.Eb+P.Eb

= 1/2×5×(100)²+0 = 0+1/2×5×(50)²+1/2×500×x²

= x = 5√3

Thus, the velocity will be 50 cm s–1 at a distance of 5√3cm