1. ⇒ (MHT CET 2023 12th May Evening Shift )

The equation of wave motion is $Y = 5 \sin (10 π t - 0.02 π x + π / 3)$ where $x$ is in metre and $t$ in second. The velocity of the wave is

A. 300 m/s

B. 400 m/s

C. 500 m/s

D. 600 m/s

Correct Option is (C)

$∴$ Equation of Wave: $y = A \sin (kx - ω t + ϕ)$

Value of $k$ :

$\begin{aligned} k = \frac{2 π}{λ} \\ k = 0.02 π \\ ω = 10 π \end{aligned}$

$∴$ Velocity of wave is $v = \frac{ω}{k}$

$\begin{aligned} v = \frac{10 π}{0.02 π} \\ v = 500 m / s \end{aligned}$

2. ⇒ (MHT CET 2023 12th May Morning Shift )

The displacement of two sinusoidal waves is given by the equation

$\begin{aligned} y_{1} = 8 \sin (20 x - 30 t) \\ y_{2} = 8 \sin (25 x - 40 t) \end{aligned}$

then the phase difference between the waves after time $t = 2 s$ and distance $x = 5 cm$ will be

A. 2 radian

B. 3 radian

C. 4 radian

D. 5 radian

Correct Option is (D)

$y_{1} = 8 \sin (20 x - 3 t)$

Substituting $x = 5 cm$ and $t = 2 s$ ,

$y_{1} = 8 \sin (40)$

Similarly, $y_{2} = 8 \sin (45)$

$∴$ phase difference $= 45 - 40 = 5$ radian

3. ⇒ (MHT CET 2023 11th May Evening Shift )

A sound of frequency $480 Hz$ is emitted from the stringed instrument. The velocity of sound in air is $320 m / s$ . After completing 180 vibrations, the distance covered by a wave is

A. 60 m

B. 90 m

C. 120 m

D. 180 m

Correct Option is (C)

Given: $v = 320 m / s, f = 480 Hz, N = 180$

$\begin{aligned} v & = f λ \\ ∴ λ & = \frac{v}{f} \end{aligned}$

Substituting the values, we get

$λ = \frac{320}{480} = \frac{2}{3}$

$∴$ The total distance covered after 180 vibrations is

$\begin{aligned} D = N \times λ \\ D = 180 \times \frac{2}{3} \\ D = 120 m \end{aligned}$

4. ⇒ (MHT CET 2023 11th May Evening Shift )

The path difference between two waves, represented by $y_{1} = a_{1} \sin (ω t - \frac{2 π x}{λ})$ and $y_{2} = a_{2} \cos (ω t - \frac{2 π x}{λ} + ϕ)$ is

A. $\frac{λ}{2 π} (ϕ)$

B. $\frac{λ}{2 π} (ϕ + \frac{π}{2})$

C. $\frac{2 π}{λ} (ϕ - \frac{π}{2})$

D. $\frac{2 π}{λ} (ϕ)$

Correct Option is (B)

$\begin{aligned} y_{1} = a_{1} \sin (ω t - \frac{2 π x}{λ}) \\ y_{2} = a_{2} \cos (ω t - \frac{2 π x}{λ} + ϕ) \end{aligned}$

$y_{2}$ can also be written as

$\begin{aligned} y_{2} & = a_{2} \sin [\frac{π}{2} + (ω t - \frac{2 π x}{λ} + ϕ)] \\ = a_{2} \sin (ω t - \frac{2 π x}{λ} + ϕ + \frac{π}{2}) \end{aligned}$

The phase difference between the two waves is

Path difference $= \frac{λ}{2 π} \times$ Phase difference

Path difference $= \frac{λ}{2 π} \times (ϕ + \frac{π}{2})$

5. ⇒ (MHT CET 2023 11th May Morning Shift )

A transverse wave $Y = 2 \sin (0.01 x + 30 t)$ moves on a stretched string from one end to another end in 0.5 second. If $x$ and $y$ are in $cm$ and $t$ in second, then the length of the string is

A. 5 m

B. 10 m

C. 15 m

D. 20 m

Correct Option is (C)

Given $Y = 2 \sin (0.01 x + 30 t)$

Comparing with standard equation,

$\begin{aligned} Y & = A \sin (kx + ω t), \\ ∴ k & = 0.01 / cm \\ ω & = 30 rad / s \end{aligned}$

Velocity $v = \frac{ω}{k} = \frac{30}{0.01} = 3000 cm / s$

$\begin{aligned} ∴ Length L & = v \times t = 30 \times 0.5 \\ = 15 m \end{aligned}$

⇒Superposition of Waves