A string is hanging from a rigid support. A transverse wave pulse is setup at the free end. The velocity v of the pulse related to the distance x covered by it is
A
v∝√x
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
v∝x
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
v∝1x
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
v∝1x2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is Av∝√x
Let the mass of the string is m and length is l.
The linear density of the string becomes ⇒ml=μ
Tension (T) at distance x from free end T=mlx=μxg ⇒Tμ=xg ⇒√Tμ=√xg ⇒v=√xg ∴v∝√x
Hence, option (a) is correct.