A wave equation which gives the displacement along the Y direction is given by y= 10^{-4} \sin (60t + 2x), where x and y are in metres and t is time in seconds. This represents a wave

  • Option 1)

    Traveling with a velocity of 30 m/s in the negative x direction

  • Option 2)

    of wavelength x metre

  • Option 3)

    Of frequency \left ( \frac{30}{\pi} \right )hertz

  • Option 4)

    All of the above

 

Answers (1)
V Vakul

As we learnt in 

Travelling Wave Equation -

y=A \sin \left ( Kx-\omega t \right )
 

- wherein

K=2\pi /\lambda

\omega = \frac{2\pi }{T}

\lambda =  wave length

T = Time period of oscillation

 

This represent a wave whose parameter are given below

\omega =  60

 \Rightarrow frequency\ f=\frac{\omega}{2\pi}=\frac{30}{\pi}\:Hz

K=2

v=\frac{\omega}{K} = 30 m/s,\:\:\:\:wavelength=\frac{2\pi}{K}=\pi

It is moving along - X direction.

All are corrrect.


Option 1)

Traveling with a velocity of 30 m/s in the negative x direction

This option is incorrect.

Option 2)

of wavelength x metre

This option is correct.

Option 3)

Of frequency \left ( \frac{30}{\pi} \right )hertz

This option is incorrect.

Option 4)

All of the above

This option is incorrect.

Preparation Products

Knockout BITSAT 2020

It is an exhaustive preparation module made exclusively for cracking BITSAT..

₹ 4999/- ₹ 1999/-
Buy Now
Knockout BITSAT 2021

It is an exhaustive preparation module made exclusively for cracking BITSAT..

₹ 4999/- ₹ 2999/-
Buy Now
Knockout BITSAT-JEE Main 2021

An exhaustive E-learning program for the complete preparation of JEE Main and Bitsat.

₹ 27999/- ₹ 16999/-
Buy Now
Knockout BITSAT-JEE Main 2020

An exhaustive E-learning program for the complete preparation of JEE Main and Bitsat.

₹ 14999/- ₹ 7999/-
Buy Now
Exams
Articles
Questions