Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

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
`        A particle mass 10 gm is executing shm with an amplitude of 0.5 m and periodic time of pi/s secThe max value of the force acting on the particle`
7 years ago

147 Points
```										Dear pallavi
x=Xsinwt         where w=2 pi/T          and T=pi/5
=.5sin 10t
w=√k/m
so k=w2m
=102 *10
=1000   gm/sec2
maximum force=kx
=1000*.5  m*gm/sec2
=.5 kg m/sec2

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.  Regards, Askiitians Experts Badiuddin

```
7 years ago
23 Points
```										Dear Pallavi,
We know for shm motion of the particle  its displacement is given by y= y0 sin ωt  Where  ω = 2π/T =  2π/π = 2 and amplitude
y0 = 0.5 m
thus aceeleration on the particle is given by a=  d2y/d2t =   - y0 * ω2 * sin ωt
Maximum acceleration is amax = - y0 * ω2  = - 2 m/s2
Thus maximum force in magnitude = | m* amax| = 0.02 N
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  !!!

Regards,
```
7 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Mechanics

View all Questions »
• Complete Physics Course - Class 12
• OFFERED PRICE: Rs. 2,756
• View Details
• Complete Physics Course - Class 11
• OFFERED PRICE: Rs. 2,968
• View Details