In shm can you please tell me the soln of this a simple pendulum of length l is suspended from a ceiling of a cart which is sliding without friction on an inclined angle theta. Determine the time period I am a bit confused why the ans is 2pie root l/gcos theta..how is it?

It is easy Here,point of suspension has an acceleration a = g sin? (down the plane).Further, g can be resolved into two components g sin? (along the plane) and g cos? (perpendicular to plane). Now, geff = g- a = g cos? T=2pie root l/geff =2pie root l/g cos?

Here `?` represents theta.

No..sry its questn mark

Ya..bt it seems to be theta

how geff= g cos theta?

a= gsin theta then why is g-gsin theta=gcos theta?

The solution u gv was frm dc Pandey example bt i ddnt understand!

My friend you are forgetting one basic principle.Acceleration is a vector quantity.You cannot directly subtract the values.Like here geff=g-a So geff=under root {g^2 + g^2(sin theta)^2 - 2*g*g sin theta*cos alpha } where alpha is the angle between g and g sin theta. If you correctly resolve g into 2 components, you will get the value of alpha = 90-theta.Using this value you will get, geff=under root {g^2 + g^2(sin theta)^2 - 2*g*g sin theta*cos (90-theta) } =under root {g^2 + g^2(sin theta)^2 - 2*g*g sin theta*sin theta } =under root {g^2 + g^2(sin theta)^2 - 2*g^2(sin theta)^2 } =under root {g^2 - g^2(sin theta)^2 } =under root {g^2(1-(sin theta)^2) } =under root {g^2(cos theta)^2 } Hence geff=g costheta

Thanx for the solution my mistake was that i assumed alpha to be 90+theta i got my mistake thanx for the help..

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