Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

A particle starts oscillating SHM from its equilibrium position than, the ratio of kinetic energy and potential energy of the particle at time T/12 is a.2:1 b.3:1 c.4:1 d.1:4

A particle starts oscillating SHM from its equilibrium position than, the ratio of kinetic energy and potential energy of the particle at time T/12 is
a.2:1
b.3:1
c.4:1
d.1:4

Grade:11

3 Answers

Gaurav Gupta
askIITians Faculty 686 Points
3 years ago
564-2293_shm.jpg
Khimraj
3007 Points
3 years ago
Total energy is given as TE = (½)mw2A2
Kinetic energy is given as KE =( ½ )mv2 =  (½)mw2A2cos2(wt).
Potential energy is given as PE = TE – KE = (½)mw2A2sin2(wt).
Value of wt = (360/T)*(T/12) = 30
So KE/PE = cot2(wt) = cot2(30) = 3.
Hope it clears.
ankit singh
askIITians Faculty 614 Points
one year ago

Let the displacement be x

KE=21K(A2x2)
PE=21Kx2
Given:
KE=PE
21K(A2x2)=21Kx2
x=2ALet the displacement be x
KE=21K(A2x2)
PE=21Kx2
Given:
KE=PE
21K(A2x2)=21Kx2
 

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free