#### Thank you for registering.

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

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
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

# plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz solve the problem 3 years ago

The time period of the pendulum in the stationary lift is T1=2 π √l/g  where l is the length of the pendulum and g is the acceleration due to gravity

When the lift start accelerating upwards

From  Newton's 2nd Law we know (ΣF=ma) acting on the pendulum. The overall acceleration of the pendulum is upward (with the lift). So ma is positive (upward). The only external forces acting on the pendulum are the force of gravity acting down (-W=-mg) and the supporting Normal Force FN upward on the pendulum

. So ΣFN=ma(net)=-ma –mk (k=acceleration of the lift)_

∴a=−(g+k)   In the problem it is given k= g/2

a=g+g/2=3g/2 ignoring the direction of the force

Let the new time period be T2 when the pendulum is accelerating with lift

T1/T2= √ [g/(3g/2)] or T2=T1√2g/3g or T2= (√2/3)T1

Hence T2 = √2 *√3 /√3 = √2 sec