Question

A body of mass 5 g is executing SHM with amplitate
10 cm, its velocity is 100 cm/s. Its velocity will
50 cm/s at a displacement from the mean position ?

Answer 1

answer

x= 5√3 cm

step by step explanation :

let the spring constant of spring be =K

Given: velocity at mean position v^m= 100cm/s

Amplitate A =10 cm

Applying conservation of energy at mean position and the extreme position

K.Eo +P.Eo= K.Ea +P.Ea

1/2mv^2m +0=0+1/2K(10)^2=

K =5dyness /cm

velocity of body is 50cm /s at point b which lies x cm away from point O

Now Applying conservation of energy at mean position and at point B

K.Eo+P.Eo=K.Eb + P.Eb

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

x=5√3cm