JEE Main Waves Practice Questions With Solutions

A plane progressive wave is given by $y=2\cos 2\pi (330\mathrm{t}-x)\mathrm{m}$y = 2 cos ⁡ 2 \pi \left(\right. 330 t - x \left.\right) m. The frequency of the wave is :

The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If length of the open pipe is $60\mathrm{cm}$60 \textrm{ } cm, the length of the closed pipe will be:

A car P travelling at $20{\mathrm{ms}}^{-1}$20 \left(\textrm{ } ms\right)^{- 1} sounds its horn at a frequency of $400\mathrm{Hz}$400 \textrm{ } Hz. Another car $\mathrm{Q}$Q is travelling behind the first car in the same direction with a velocity $40{\mathrm{ms}}^{-1}$40 \left(\textrm{ } ms\right)^{- 1}.The frequency heard by the passenger of the car $\mathrm{Q}$Q is approximately [Take, velocity of sound $=360{\mathrm{ms}}^{-1}$= 360 \left(\textrm{ } ms\right)^{- 1} ]

For a periodic motion represented by the equation

$y=\sin \omega \mathrm{t}+\cos \omega \mathrm{t}$y = sin ⁡ \omega t + cos ⁡ \omega t

the amplitude of the motion is

The engine of a train moving with speed $10{\mathrm{ms}}^{-1}$10 \left(\textrm{ } ms\right)^{- 1} towards a platform sounds a whistle at frequency $400\mathrm{Hz}$400 \textrm{ } Hz. The frequency heard by a passenger inside the train is: (neglect air speed. Speed of sound in air $=330{\mathrm{ms}}^{-1}$= 330 \left(\textrm{ } ms\right)^{- 1} )

A steel wire with mass per unit length $7.0\times {10}^{-3}\mathrm{kg}{\mathrm{m}}^{-1}$7.0 \times 10^{- 3} \textrm{ } kg \left(\textrm{ } m\right)^{- 1} is under tension of $70\mathrm{N}$70 \textrm{ } N. The speed of transverse waves in the wire will be:

A person observes two moving trains, 'A' reaching the station and 'B' leaving the station with equal speed of $30\mathrm{m}/\mathrm{s}$30 \textrm{ } m / s. If both trains emit sounds with frequency $300\mathrm{Hz}$300 \textrm{ } Hz, (Speed of sound: $330\mathrm{m}/\mathrm{s}$330 \textrm{ } m / s) approximate difference of frequencies heard by the person will be:

A travelling wave is described by the equation

$y(x,t)=[0.05\sin (8x-4t)]$y \left(\right. x , t \left.\right) = \left[\right. 0.05 sin ⁡ \left(\right. 8 x - 4 t \left.\right) \left]\right. m

The velocity of the wave is : [all the quantities are in SI unit]

In the wave equation

$y=0.5\sin \frac{2\pi }{\lambda }(400\mathrm{t}-x)\mathrm{m}$y = 0.5 sin ⁡ \frac{2 \pi}{\lambda} \left(\right. 400 t - x \left.\right) m

the velocity of the wave will be:

A transverse wave is represented by $y=2\sin (\omega t-kx)\mathrm{cm}$y = 2 sin ⁡ \left(\right. \omega t - k x \left.\right) cm. The value of wavelength (in $\mathrm{cm}$cm) for which the wave velocity becomes equal to the maximum particle velocity, will be :