a simple pendulum with bob of mass m and length x is held in position at an angle theta1 and theta2 with vertical.When released from these positions,speeds with which it passes the lowest positions are v1 and v2 respectively.Then v1÷v2 is ?
Answers
Answered by
79
KE = loss in PE
m v1^2 /2 = m g L (1 - cos θ1)
So v1 = sqrt(g*L) * 2* Sin (θ1 /2)
Similarly v2 = sqrt(g L)* 2 sin(θ2 /2).
Ratio v1 / v2 = Sin (θ1 /2) ÷ Sin(θ2 /2).
m v1^2 /2 = m g L (1 - cos θ1)
So v1 = sqrt(g*L) * 2* Sin (θ1 /2)
Similarly v2 = sqrt(g L)* 2 sin(θ2 /2).
Ratio v1 / v2 = Sin (θ1 /2) ÷ Sin(θ2 /2).
kvnmurty:
Click on the red hearts thanks
Answered by
9
The solution is provided in the attachment.
PLEASE MARK AS BRAINLIEST
PLEASE MARK AS BRAINLIEST
Attachments:
Similar questions