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
`        the position of a particle moving along x-axis varies with time t according to equation- x=under root 3 sin omege t- cos omega t where omega is constant.find the region in which the particle is confined.`
2 years ago

Nicho priyatham
626 Points
```										by solving above  equationx=2sin{(omega)t-30}so the partical can travel from -2to+2 on x axisso range is [-2,+2]PLZ APPROVE if useful n in case of any doubts put it in coment box
```
2 years ago
vindhya
50 Points
```										i’m sorry, but i am unable to understand. So, if you don't mind, please explain again. Thanks
```
2 years ago
erra akhil
208 Points
```										Dear Vidhya,Given, x=(root 3 )sin omega t- cos omega tIn trigonometry we knw that,maximum of {a sin theta + b cos theta}=root over{a2+b2}minimum of {a sin theta + b cos theta}= – root over{a2+b2}Similarly in your question,maximum value of x=root over{ (root 3)2-(1)2 };minimum value of x= – root over{(root 3)2+(1)2};So, x lies between -2 and +2.Region in which particle is confined = [ - 2 , + 2 ]Approve if my answer helped you and in case any doubts put it in the comment box.
```
2 years ago
vindhya
50 Points
```										thanks a lot erra akhil sir, but my name is vindhya and not vidhya.
```
2 years ago
erra akhil
208 Points
```										okay Vindhya.
```
2 years ago
Think You Can Provide A Better Answer ?

Other Related Questions on General Physics

View all Questions »
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details
• Kinematics & Rotational Motion
• OFFERED PRICE: Rs. 636
• View Details