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

Total Price: Rs.

There are no items in this cart.
Continue Shopping
pallavi pradeep bhardwaj Grade: 12

A mass m=100gm is attached at the end of  a light spring which oscillates on a fictionless horizontal table with an amplitude equal to 0.16m and time period equal to 2sec Initially the mass is released from rest at t=0 and displacement x=-0.16m the expression for the displacement of the mass at any time t is

8 years ago

Answers : (1)

Badiuddin askIITians.ismu Expert
147 Points

Dear Pallavi

Equation of SHM is given as

x=X sin(wt + Φ)   

  where  X=.16 m




so x=.16 sin(∏t + Φ)  

given at t=0 sec   x=-.16m

-.16= .16sin(0 + Φ) 

or sin( Φ) = -1

   or  Φ =-∏/2

so final equation is

x=.16 sin(∏t -∏/2)

Please feel free to post as many doubts on our discussion forum as you can.
 If you find any question Difficult to understand - post it here and we will get you the answer and detailed

solution very quickly.
 We are all IITians and here to help you in your IIT JEE  & AIEEE preparation.

 All the best.
Askiitians Experts

8 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete Physics Course - Class 12
  • OFFERED PRICE: Rs. 2,756
  • View Details
Get extra Rs. 220 off
USE CODE: Neerajwin
  • Complete Physics Course - Class 11
  • OFFERED PRICE: Rs. 2,968
  • View Details
Get extra Rs. 237 off
USE CODE: Neerajwin

Ask Experts

Have any Question? Ask Experts

Post Question

Answer ‘n’ Earn
Attractive Gift
To Win!!! Click Here for details