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 pendulum bob of mass m hangs from a massles elastic string. The potential energy (elastic+gravitational) of the system (bob+string+ Earth) measured realtive to the position of the of the bob corresponding to the normal length of the string is(where x= static deformation (elongation) of the string)

A pendulum bob of mass m hangs from a massles elastic string. The potential energy (elastic+gravitational) of the system (bob+string+ Earth) measured realtive to the position of the of the bob corresponding to the normal length of the string is(where x= static deformation (elongation) of the string)

Grade:Upto college level

1 Answers

Navneet kumar
37 Points
7 years ago
Let the normal length of the string = L Let the bob hang from a height = h Elastic PE of the string = 0 Gravitational potential of the mass = mgh Total potential Ui = mgh Now after the string is elongated by length x elastic potential energy = ½kx2 Gravitational potential energy = mg(h – x) Total potential energy Uf = mg(h – x) + ½kx2 Again, F = -kx F = mg => k = mg/x in magnitude. So, it can be neglected from denominator => ?U = Uf – Ui = [mg(h – x) + ½kx2] – mgh = -mgx + ½mgx => ?U = - ½mgx

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