Question

A particle is exicuting SHM with amplitude A and has maximum velocity v0.its speed at displace ment 3A/4 will be

Answer 1

v = ω√(a^2 – y^2)

maximum speed is v0 = ωA-----(i)

then 
v = ω√[A^2 - (3A/4)^2]
v = ω√[A^2 - 9A^/16]
v = ω√[7A^2/16]
v = ωA x (√7)/4
v= v0 [√7/4]


i hope it will help you
regards