MY CART (5)

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

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: R

There are no items in this cart.
Continue Shopping
Menu
sudeep joseph Grade: 11
        

a particle of mass 0.50 kg executes a simple harmonic motion under a for F = -(50N/m)x. if it crosses the centre of oscillation with a speed of 10m/s find the amplitude of the motion.

6 years ago

Answers : (1)

Askiitians Expert Soumyajit IIT-Kharagpur
28 Points
										

Dear Sudeep Joseph,


Ans:- The energy at the mean position is totally kinetic and at max amplitude it is totally potential now the potential energy may be calculated like this work done in moving a small distance dx against the forceis  = -F dx=50 x dx leet A be the amplitude then integrating the function within 0 to A we get, 1/2(50 A²) is the PE now it should be equal to the KE as the total energy is conserved in SHM. So, solving we get (0.5)(10)²=50 A²


or A=1 meter(ans)


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 preparation.



All the best Sudeep Joseph !!!


 



Regards,


Askiitians Experts
Soumyajit Das IIT Kharagpur

6 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete Physics Course - Class 12
  • OFFERED PRICE: R 2,600
  • View Details
  • Complete Physics Course - Class 11
  • OFFERED PRICE: R 2,800
  • View Details

Ask Experts

Have any Question? Ask Experts

Post Question

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