 ×     #### 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
```
33. A wave is described by the equation y = (1.0 mm) sin pi t  2.0cm 0.01s (a) Find the time period and the wavelength. (b) Write the equation for the velocity of the particles. Find the speed of the particle at x = 1.0 cm at time t = 0.01 s. (c) What are the speeds of the particles at x = 3.0 cm, 5.0 cm and 7.0 cm at t = 0.01 s? (d) What are the speeds of the particles at x = 1.0 cm at t = 0.011, 0.012, and 0.013 s?

```
3 years ago Saurabh Koranglekar
10341 Points
```							Dear studentPlease write the Question in a standard form or attach an image of the questionRegards
```
8 months ago
```							Y = (1 mm) sin π [x/2cm – t/0.01 sec] a) T = 2  0.01 = 0.02 sec = 20 ms λ = 2 * 2 = 4 cm b) v = dy/dt = d/dt [sin 2π (x/4 – t/0.02)] = –cos2π {x/4) – (t/0.02)} * 1/(0.02) ⇒ v = –50 cos 2π {(x/4) – (t/0.02)} at x = 1 and t = 0.01 sec, v = –50 cos 2* [(1/4) – (1/2)] = 0 c) i) at x = 3 cm, t = 0.01 sec v = –50 cos 2π (3/4 – ½) = 0 ii) at x = 5 cm, t = 0.01 sec, v = 0 (putting the values) iii) at x = 7 cm, t = 0.01 sec, v = 0 at x = 1 cm and t = 0.011 sec v = –50 cos 2π {(1/4) – (0.011/0.02)} = –50 cos (3/5) = –9.7 cm/sec (similarly the other two can be calculated)
```
8 months ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Wave Optics

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