We have given, y=0.03sin(450t−9x) Comparing it with standard equation of wave, we get ω= 450k=9 ∴v=kω=9450=50 m/s Velocity of travelling wave on a stretched string is given by v=μT⇒μT=2500 μ= linear mass density ⇒T=2500×5×10−3 ⇒12.5 N
Equation of travelling wave on a stretched string of linear density 5 ~g / m is y=0.03 sin (450 t-9 x) where distance and time are measured in SI…
JEE Main 2019 — Physics Waves & Oscillations
2019mcqmedium
Equation of travelling wave on a stretched string of linear density 5 g/m is y=0.03sin(450t−9x) where distance and time are measured in SI units. The tension in the string is:
Official previous-year question
Held on 11 Jan 2019 · Verified 6 Jul 2026.
Options
- A
10 N
- B
7.5 N
- C
12.5 N
- D
5 N
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