badge image

Enroll For Free Now & Improve Your Performance.

×
User Icon
User Icon
User Icon
User Icon
User Icon

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
Menu
Grade: 12th pass

                        

A particle starts SHM from mean position and has time period The. The time taken by it to complete 15/8 oscillation is

2 years ago

Answers : (1)

Arun
24742 Points
							
 

Total distance covered by the particle = 4A.

 

we divide this whole path in 8 intervals of A/2.

 

so, 5/8 oscillations means, it has already completed 1/2 oscillation(i.e. total dist. = 2A) and is half way to the other side i.e. A/2.

 

so, A/2 = Asinwt, w= 2pi/ T ,

 

substitute to get t= T/12.

 

now, total tym taken = tym to complete previous one half(2A) + tyn taken to complete A/2 = t/2 + t/12 = 7T/12

2 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 731 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 18 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts

Post Question

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