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: 11

                        

A particle doing simple harmonic motion whose amplitude is 4cm,time period is 12 sec.the ratio between time taken by it in going from its mean position to 2 cm and 2 cm to extreme position is

3 years ago

Answers : (1)

Arun
24742 Points
							
y = 4 sin(2 pi/12)*t
Now y = 2
2 = 4 sin(pi/6) *t1
1/2 = sin(pi/6) *t1
Pi/6 = (pi/6)* t1
t1 = 1 sec
Hence T1 = t1 - 0 = 1 sec
 
Now, similarily
Put y = 4
You will get t2 = 3 sec.
Hence T2 = t2 - t1 = 3- 1 = 2 sec
Hence
T1 : T2 = 1:2
 
3 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

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


Course Features

  • 110 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