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 of mass 0.1kg is executing SHM of amplitude 1m and period 0.2 sec what max. Value of force acting on it? If oscillation is produced by spring what is the force constant of spring
7 months ago

Answers : (1)

Arun
13086 Points
							
Dear student
 
Maximum acceleration
a(max) = w² *a = 4\pi² A/ T² = 4 *(3.14)² * 1/(0.2)²
 
F(max) = m * a(max) = 0.1 * 4 * (3.14)²*1/(0.2)²
 = 98.6 N
 
Force constant k = Fmax/A = 98.6/1
 = 98.6 N/m
 
 
Regards
Arun (askIITians forum expert)
 
7 months ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Get Extra Rs. 276 off

COUPON CODE: SELF10


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


Get Extra Rs. 297 off

COUPON CODE: SELF10

Ask Experts

Have any Question? Ask Experts

Post Question

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