7:26 AM
@JohnRennie hi sir can u guess a curve whose tangential and radial acc are equal

7:58 AM
@JackRod I don't know I'm afraid. Sorry :-(

8:29 AM
@JohnRennie hi

Hi :-)

sir I have a question, A point P describes an equiangular spiral $r=ae^{\theta}cot\alpha$
with constant angular velocity about the pole O
show that acc varies as OP and it is in direction making with tangent at P the same angle that OP makes
OP is a spircal curve

I have no idea I'm afraid. Maths isn't my strong point.

just can u make me understand how the things is happening
in the solution they have made a triangle with a string attached to O spiraling around till P
P having angle $\psi$
@JohnRennie

What is α in the equation?

8:42 AM
a constant
@JohnRennie

So the curve is just r = A exp(θ) where A is the constant a cotα ?

no sir i though your were asking aboutA=a $\alpha$ s another angle which equal to $\psi$
** silly english error no sir I thought u were asking about A=a ,hefre $\alpha$ is equal to $\psi$

I'm getting confused now. Can you post a picture of the explanation in the book?

@JohnRennie

9:14 AM
@JohnRennie leave that sir
one last question a particle performs shm in t=1 it travel a
and t=2 it travels b
I need to find amplitude and time period
I m gettinf A=2a^2/3a-b

Do you mean y(1) = a and y(2) = b?

10:07 AM
yes
@JohnRennie

11:05 AM
If we write y = A sin(ωt) then from y(1) = a and y(2) = b we get:
A sin(ω) = a
A sin(2ω) = b
I guess we can write sin(2ω) = sin(ω)cos(ω) = sin(ω) √1 - sin²(ω)
So the second equation becomes:
A sin(ω) √1 - sin²(ω) = b
Then the first equation gives us sin(ω) = a/A and we can substitute to get:
A a/A √1 - a²/A² = b
And then you can solve for A
Oops, missed a factor of 2. The second equation should be:
2A sin(ω) √1 - sin²(ω) = b
So we get:
2A a/A √1 - a²/A² = b
I get:
A² = 4a⁴/(4a² - b²)