A sinusoidal wave of wavelength 7.5 cm travels a distance of 1.2 cm along the x -direction in 0.3 sec . The crest P is at x=0 at t=0 sec and maximum displacement of the wave is 2 cm . Which equation correctly represents this wave?

Waves PYQ — JEE Main 2025 · 2 Apr · Shift 2 · Q55 with Solution

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Correct answer: A

Given data

Wavelength: $\lambda = 7.5\,\text{cm}$
Distance travelled in $0.3\,\text{s}$ : $1.2\,\text{cm}$
Amplitude: $A = 2\,\text{cm}$
At $t=0$ , a crest is at $x=0$

So the wave speed is

v = \frac{1.2}{0.3} = 4\,\text{cm/s}

Find wave number $k$

For a sinusoidal wave,

k = \frac{2\pi}{\lambda} = \frac{2\pi}{7.5}

Numerically,

k \approx \frac{6.283}{7.5} \approx 0.84\,\text{rad/cm}

This matches approximately with $0.83$ used in the options.

Find angular frequency $\omega$

Using

\omega = vk

we get

\omega = 4 \times 0.84 \approx 3.35\,\text{rad/s}

Choose correct wave form

A wave travelling in the positive $x$ -direction is written as

y = A \cos(kx-\omega t+\phi)

or equivalently with sine.

Since at $t=0$ a crest is at $x=0$ , we need maximum displacement there:

y(0,0)=+A=2

This is satisfied by choosing phase constant $\phi=0$ in cosine form:

y = 2\cos(kx-\omega t)

Thus,

y = 2\cos(0.83x - 3.35t)\,\text{cm}

Check options

A: $y=2\cos(0.83x-3.35t)$ ✅ Matches amplitude, $k$ , $\omega$ , and crest condition.
B: $y=2\sin(0.83x-3.5t)$ ❌ At $x=0,t=0$ , $y=0$ , not a crest.
C: $y=2\cos(0.13x-0.5t)$ ❌ Wrong $k$ and $\omega$ .
D: $y=2\cos(3.35x-0.83t)$ ❌ $k$ and $\omega$ are interchanged.

Therefore the correct option is A.

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View written solutionFree

Correct answer: A

Given data

Wavelength: $\lambda = 7.5\,\text{cm}$
Distance travelled in $0.3\,\text{s}$ : $1.2\,\text{cm}$
Amplitude: $A = 2\,\text{cm}$
At $t=0$ , a crest is at $x=0$

So the wave speed is

v = \frac{1.2}{0.3} = 4\,\text{cm/s}

Find wave number $k$

For a sinusoidal wave,

k = \frac{2\pi}{\lambda} = \frac{2\pi}{7.5}

Numerically,

k \approx \frac{6.283}{7.5} \approx 0.84\,\text{rad/cm}

This matches approximately with $0.83$ used in the options.

Find angular frequency $\omega$

Using

\omega = vk

we get

\omega = 4 \times 0.84 \approx 3.35\,\text{rad/s}

Choose correct wave form

A wave travelling in the positive $x$ -direction is written as

y = A \cos(kx-\omega t+\phi)

or equivalently with sine.

Since at $t=0$ a crest is at $x=0$ , we need maximum displacement there:

y(0,0)=+A=2

This is satisfied by choosing phase constant $\phi=0$ in cosine form:

y = 2\cos(kx-\omega t)

Thus,

y = 2\cos(0.83x - 3.35t)\,\text{cm}

Check options

A: $y=2\cos(0.83x-3.35t)$ ✅ Matches amplitude, $k$ , $\omega$ , and crest condition.
B: $y=2\sin(0.83x-3.5t)$ ❌ At $x=0,t=0$ , $y=0$ , not a crest.
C: $y=2\cos(0.13x-0.5t)$ ❌ Wrong $k$ and $\omega$ .
D: $y=2\cos(3.35x-0.83t)$ ❌ $k$ and $\omega$ are interchanged.

Therefore the correct option is A.

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