 ×     #### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+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
```
The potential energy of a harmonic oscillator of mass 2 kg in its mean position is 5 J. If its total energy is 9 J and its amplitude is 0.01 m, its time period will beI have no understood that why have they found out KE and the equated it to its formula Where as we have got formula for TE and we even know it`s formula Pls.explain This is the link of its soultion I got on your website http://www.askiitians.com/iit-study-material/iit-jee-physics/oscillation/energy-in-simple-harmonic-motion/

```
2 years ago

```							Maximum velocity occurs at mean position in the simple harmonic motionK.E at the centre= 9-5 = 4J=>1/2mv^2 = 4 => V(max) =2m/sIn SHM V(max) = AW where A=1 in the given problem =>W=2 =>f=3.14  (f=2*3.14/W)
```
2 years ago 605 Points
```							Dear student,Maximum velocity occurs at mean position in the simple harmonic motionK.E at the centre= 9-5 = 4J=>1/2mv^2 = 4 => V(max) =2m/sIn SHM V(max) = AW where A=1 in the given problem =>W=2 =>f=3.14  (f=2*3.14/W) Thanks and regards,Kushagra
```
3 months ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Mechanics

View all Questions »  ### 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