×

#### 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
```
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

## Other Related Questions on General Physics

View all Questions »

### 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