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```
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?

```
6 years ago

```							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?
```
6 years ago
```							Here `?` represents theta.
```
6 years ago
```							No..sry its questn mark
```
6 years ago
```							Ya..bt it seems to be theta
```
6 years ago
```							how geff= g cos theta?
```
6 years ago
```							a= gsin theta then why is g-gsin theta=gcos theta?
```
6 years ago
```							The solution u gv was frm dc Pandey example bt i ddnt understand!
```
6 years ago
```							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
```
6 years ago
```							Thanx for the solution my mistake was that i assumed alpha to be 90+theta i got my mistake thanx for the help..
```
6 years ago
```							Moral: never solve physics late night it can make u do wonders!!
```
6 years ago
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