A standing wave ξ = a sin kx.cos ωt is maintained in a homogeneous rod with cross-sectional area S and density ρ . If the total mechanical-energy confined between the sections corresponding to the adjacent displacement nodes is 
πS ρ ω 2 a 2 /k Find p.
Text Solution
Verified by ExpertsCHECK THE SOLUTION
(4)
If we consider AB part of rod then every particle in AB is doing S.H.M. if we determine max kinetic energy of S.H.M. of each particle which is equal to mechanical energy of vibration and sum of all these energy between these (adjacent displacement node) gives mechanical energy confined between it kinetic energy of any particle of mass ( ρ sdx) = 
ρ sdx 
.
d KE = 
ρ s dx ω 2 a 2 sin 2 kx sin 2 ω t
d KEmax = 
ρ s ω 2 a 2 sin 2 kx dx
net energy (mechanical) between A to B (0 to λ /2)
= 
ρ s ω 2 a 2 
= 
ρ s ω 2 a 2 
= 
ρ s ω 2 a 2 
= 
πρ s ω 2 
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems