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
Amit Saxena Grade: upto college level
`        The maximum speed and acceleration of a particle executing simple harmonic motion are 10 cm s-1 and 50 cm s-2. Find the position(s) of the particle when the speed is 8 cm s-1.`
3 years ago

Navjyot Kalra
654 Points
```										Sol. vmax = 10 cm/sec.
⇒ rω = 10
⇒ ω2 = 100/r2  ….(1)
Amax = ω2r = 50 cm/sec
⇒ ω2 = 50/y = 50/r    …..(2)
∴ 100/r2 = 50/r ⇒ r = 2 cm.
∴ ω = √(100/r^2 ) = 5 sec2
Again, to find out the positions where the speed is 8m/sec,
v2 = ω2 (r2 – y2)
⇒ 64 = 25 ( 4 – y2)
⇒ 4 – y2 = 64/25 ⇒ y2 = 1.44 ⇒ y = √1.44 ⇒ y = ± 1.2 cm from mean position.

```
3 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
Get extra Rs. 413 off
USE CODE: COUPON10
• Complete Physics Course - Class 11
• OFFERED PRICE: Rs. 2,968
• View Details
Get extra Rs. 445 off
USE CODE: COUPON10