Equation of travelling wave on stretched string of linear density 5g/m is y = 0.003\sin \left ( 450t - 9x \right ) where distance and time are measured in SI units. The tension in the string is:

  • Option 1)

    5N

  • Option 2)

    7.5N

  • Option 3)

    10N

  • Option 4)

    12.5 N

Answers (1)
A admin

 

Relation between particle velocity and wave speed -

V_{P}= -V\: \frac{dy}{dx}

- wherein

V_{P}= particle velocity

V = wave velocity

\frac{dy}{dx}= slope of curve

 

 

Speed of wave on string -

v= \sqrt{\frac{T}{\mu }}
 

- wherein

T= Tension in the string

\mu = linear mass density

 

V=\frac{\omega }{K}=\frac{450}{9}=50m/s

V=\sqrt{\frac{T}{M}}

\Rightarrow \frac{T}{M}=2500

T=2500\times 5\times 10^{-3}

=12.5 N

 

 

 

 


Option 1)

5N

Option 2)

7.5N

Option 3)

10N

Option 4)

12.5 N

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