1. (JEE Main 2023 (Online) 24th January Morning Shift)

A travelling wave is described by the equation

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

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

A. $4 {ms}^{- 1}$

B. $2 {ms}^{- 1}$

C. $8 {ms}^{- 1}$

D. $0.5 {ms}^{- 1}$

Correct Answer is Option (D)

$∵ y (x, t) = [0.05 \sin (8 x - 4 t)] m$

$\begin{aligned} Speed of wave & = | \frac{Coefficient of t}{Coefficient of x} | \\ = \frac{4}{8} = 0.5 {ms}^{- 1} \end{aligned}$

2. (JEE Main 2023 (Online) 11th April Morning Shift)

The equation of wave is given by

$Y = 10^{- 2} \sin 2 π (160 t - 0.5 x + π / 4)$

where $x$ and $Y$ are in $m$ and $t$ in $s$ . The speed of the wave is ________ $km h^{- 1}$ .

Correct Answer is 1152

Given the wave equation:

$Y = 10^{- 2} \sin 2 π (160 t - 0.5 x + π / 4)$

Comparing this equation with the general form:

$Y = A \sin (2 π (f t - k x + ϕ))$

We can identify the wave number $k = 0.5 m^{- 1}$ and the frequency $f = 160 Hz$ . The wave speed $v$ can be found using the relationship between wave number, wave speed, and frequency:

$v = \frac{ω}{k} = \frac{2 π f}{2 π k}$

Now, we can calculate the wave speed:

$v = \frac{2 π \times 160}{2 π \times 0.5} = \frac{160}{0.5} m / s$

$v = 320 m / s$

Now, we need to convert the wave speed from meters per second to kilometers per hour:

$v = 320 \frac{m}{s} \times \frac{1 km}{1000 m} \times \frac{3600 s}{1 h}$

$v = 320 \times \frac{1}{1000} \times 3600 km / h$

$v = 1152 km / h$

So, the speed of the wave is $1152 km / h$ .

3. (JEE Main 2023 (Online) 10th April Morning Shift)

A transverse harmonic wave on a string is given by

$y (x, t) = 5 \sin (6 t + 0.003 x)$

where x and y are in cm and t in sec. The wave velocity is _______________ ms $^{- 1}$ .

Correct Answer is 20

The general equation for a transverse harmonic wave on a string is given by:

$y (x, t) = A \sin (k x - ω t + ϕ)$

where $A$ is the amplitude of the wave, $k$ is the wave number, $ω$ is the angular frequency, and $ϕ$ is the phase constant. The wave velocity $v$ is related to the wave number and angular frequency by the formula:

$v = \frac{ω}{k}$

Comparing the given equation with the general equation, we can see that:

$A = 5 cm$

$k = 0.003 {cm}^{- 1}$

$ω = 6 rad/s$

Therefore, the wave velocity is:

$v = \frac{ω}{k} = \frac{6}{0.003} = 2000 cm/s = 20 m/s$

4. (JEE Main 2023 (Online) 25th January Morning Shift)

The distance between two consecutive points with phase difference of 60 $^{\circ}$ in a wave of frequency 500 Hz is 6.0 m. The velocity with which wave is travelling is __________ km/s

Correct Answer is 18

JEE Main 2023 (Online) 25th January Morning Shift Physics - Waves Question 13 English Explanation

$\begin{aligned} Δ x = \frac{λ}{2 π} \times (\frac{π}{3}) = (\frac{λ}{6}) \\ \Rightarrow \frac{λ}{6} = 6 m \\ λ = 36 m \\ U = f λ = 500 Hz \times 36 \\ = 18000 m / s \\ = 18 km / s \end{aligned}$

5. (JEE Main 2022 (Online) 28th July Morning Shift)

In the wave equation

$y = 0.5 \sin \frac{2 π}{λ} (400 t - x) m$

the velocity of the wave will be:

A. 200 m/s

B. 200 $\sqrt{2}$ m/s

C. 400 m/s

D. 400 $\sqrt{2}$ m/s

Correct Answer is Option (C)

$v_{w a v e} = | \frac{c o e f f i c i e n t o f t}{c o e f f i c i e n t o f x} |$

$= \frac{400}{1} = 400$ m/s