Wave motionMedium
Question
A standing wave is created on a string of length 120 m and it is vibrating in 6th harmonic. Maximum possible amplitude of any particle is 10 cm and maximum possible velocity will be 10 cm/s. Choose the correct statement.
Options
A.Angular wave number of two waves will be 
.
.B.Time period of any particle’s SHM will be 4π sec.
C.Any particle will have same kinetic energy as potential energy.
D.Amplitude of interfering waves are 10 cm each.
Solution
= 120 ⇒ λ = 40 ⇒ k = 
⇒ Aω = vmax ⇒ ω = 1 ⇒ T = 2πCreate a free account to view solution
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