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
        Stationery waves produced on a stretch string which is 1.5 m long and fixed at its ends if six loops are formed find them wavelength and the distance between the nodes and adjacent antinodes
11 months ago

Answers : (1)

Eshan
askIITians Faculty
1987 Points
							Dear student,

Each loop aquires a length of half the wavelength and the total of 6 loops must be equal to the total length of the string.

Hence6\times \dfrac{\lambda}{2}=1.5m\implies \lambda=0.5m
Distance between node and antinode is half the length of each loop

=\dfrac{1}{2}\times\dfrac{\lambda}{2}=\dfrac{\lambda}{4}=0.125m
6 months 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