Verified Solution • 2025–26 Session moderate

QJEEM19Ph6598: JEE Main 2019 Physics - Class 11 JEEM19Ph6598 Solution: Equation of travelling wave on a stretched string of linear…

Equation of travelling wave on a stretched string of linear density 5 g/m is y = 0.03 sin(450 t – 9x) where distance and time are measured in SI units. The tension in the string is :

Multiple Choice Options

  • A 10 N
  • B 7.5 N
  • C 5 N
  • D 12.5 N

Step-by-Step Verified Solution

y = 0.03 sin(450 t −- 9x)

v = ωk=4509{\omega \over k} = {{450} \over 9} = 50m/s

v = Tμ⇒Tμ\sqrt {{T \over \mu }} \Rightarrow {T \over \mu } = 2500

⇒\Rightarrow   T = 2500 ×\times 5 ×\times 10 −- 3

= 12.5 N

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring