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
Menu
Grade: 12

                        

The equation of a travelling sound wave is y = 6.0 sin (600 t – 1.8 x) where y is measured in 10–5 m, t in second and x in metre. (a) Find the ratio of the displacement amplitude of the particles to the wavelength of the wave. (b) Find the ratio of the velocity amplitude of the particles to the wave speed.

3 years ago

Answers : (1)

Arun
24736 Points
							
Here given r base y = 6.0 * 10^-5 m a) Given 2π/λ = 1.8 ⇒ λ = (2π/1.8) So, r base y/ π = 6.0 * (1.8) * 10^-5 m/s /2π = 1.7 * 10^-5 m b) Let, velocity amplitude = V base y V = dy/dt = 3600 cos (600 t – 1.8) × 10^–5 m/s Here V base y = 3600 × 10^–5 m/s Again, λ = 2π/1.8 and T = 2π/600 ⇒ wave speed = v = λ/T = 600/1.8 = 1000 / 3 m/s. So the ratio of (V base y/v) = 3600 * 3 *10^-5/1000 
2 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


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


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details