Use Coupon: CART20 and get 20% off on all online Study Material

Total Price: Rs.

There are no items in this cart.
Continue Shopping
Grade: 11
A loudspeaker produces a musical sound by the oscillation of a diaphragm. If the amplitude of oscillation is limited to 1.20 × 10-3 mm, what frequencies will result in the acceleration of the diaphragm exceeding g?
4 years ago

Answers : (1)

Aditi Chauhan
askIITians Faculty
396 Points
							Here the value of maximum acceleration ax of the diaphragm is equal to the acceleration due to gravity g and which has value 9.81 m/s2 on the surface of Earth.
To obtain the frequency of diaphragm, substitute 9.81 m/s2 for acceleration ax, 1.20×10-3 mm for the amplitude of oscillation in the equation f = (√ ax/ xm) /2π,
f = (√ ax/ xm) /2π
= (√(9.81 m/s2 /1.20×10-3 mm) (1 mm/10-3 m )) /2×3.14
= (455/s) (1 Hz/(1/sec))
= 455 Hz
From the above observation we conclude that the frequency of diaphragm would be 455 Hz.
4 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
To Win!!! Click Here for details