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: 12th pass
        
the length of the wire shown in figure between the pulley is 1.5 m and its mass is 12 g. Find the frequency of vibration with which the wire vibrates in two loops leaving the midlle point of the wire between the pulley rest.
2 years ago

Answers : (1)

Dewansh
19 Points
							Tension T= mg  That is T= 9×9.8 = 88.2 N . mass per unit length of the string can be calculated,   this is equal to 0.8 × 10^ -2 we know velocity of wave = √{tension/mass per unit length}So wave velocity is 105m/sIts given that there are 2 loops and the entire wire is vibrating except the mid point so it becomes a node. And 2loops , one up second down on either side of mid point ,  A wavelength. wavelength = length of wire 1.5 mWe know wavelength , velocity So frequency can be calculated by v=∆f where ∆ = wavelength Frequency is 70 Hz.
						
one year ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Get Extra Rs. 276 off

COUPON CODE: SELF10


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


Get Extra Rs. 297 off

COUPON CODE: SELF10

Ask Experts

Have any Question? Ask Experts

Post Question

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