Question

At the mean position the the kinetic energy of a particle in SHM is :

(A) 1/6 m (B) 6 m (C) 1/36 m (D) 1/√6 m​

Answer 1

Since a SHM can be represented by x=Asin(ωt+ϕ)

Since a SHM can be represented by x=Asin(ωt+ϕ)⇒x=Asin((T2π)t+ϕ) ; T=6s

Since a SHM can be represented by x=Asin(ωt+ϕ)⇒x=Asin((T2π)t+ϕ) ; T=6sso the equation becomes 

Since a SHM can be represented by x=Asin(ωt+ϕ)⇒x=Asin((T2π)t+ϕ) ; T=6sso the equation becomes x=Asin(3πt+ϕ)

Since a SHM can be represented by x=Asin(ωt+ϕ)⇒x=Asin((T2π)t+ϕ) ; T=6sso the equation becomes x=Asin(3πt+ϕ)We are given that at t=1s,x=0; so we have

Since a SHM can be represented by x=Asin(ωt+ϕ)⇒x=Asin((T2π)t+ϕ) ; T=6sso the equation becomes x=Asin(3πt+ϕ)We are given that at t=1s,x=0; so we have0=Asin(3π+ϕ)⇒(3π+ϕ)=0⇒ϕ=−3π⇒x=Asin(3πt−3π)⇒v=dtdx=dtd(Asin(3πt−3π))=3πAcos(3πt−3π)

Since a SHM can be represented by x=Asin(ωt+ϕ)⇒x=Asin((T2π)t+ϕ) ; T=6sso the equation becomes x=Asin(3πt+ϕ)We are given that at t=1s,x=0; so we have0=Asin(3π+ϕ)⇒(3π+ϕ)=0⇒ϕ=−3π⇒x=Asin(3πt−3π)⇒v=dtdx=dtd(Asin(3πt−3π))=3πAcos(3πt−3π)Now we are given that at t=2s,∣v∣=