Oscillations and Waves NEET 2026: Phase Difference

What is the answer to the Oscillations and Waves NEET 2026 phase-difference question?

Option B is correct for the Oscillations and Waves NEET 2026 question: convert the separation from metres to centimetres before multiplying by the wave number. The two points are compared at the same instant, so the shared time-dependent term cancels.

A travelling harmonic disturbance follows this equation, with position and displacement measured in centimetres and time in seconds:

y (x, t) = 2.0 \cos [2 π (10 t - 0.0080 x + 0.35)] .

Find the phase-difference magnitude between the oscillations at two points separated by 0.5 m. The supplied choices are:

This is a NEET 2026 Physics single-correct question, tagged medium on this question bank’s scale. Its bank target time is 60 seconds, not a measured student completion time.

How do you read the complete phase and identify the wave number?

The phase is the entire argument of the cosine, including the factor outside the bracket. Following the official method, write the full phase before identifying the wave number:

ϕ (x, t) = 2 π (10 t - 0.0080 x + 0.35) .

Expanding gives:

ϕ (x, t) = 20 π t - 2 π \times 8 \times 10^{- 3} x + 0.70 π .

Keep the time variable in the first term. Omitting it incorrectly removes the phase’s dependence on time.

Compare with the standard form: $ϕ = ω t - k x + ϕ_{0} .$

The positive wave-number magnitude is:

k = 2 π \times 8 \times 10^{- 3} = 0.016 π r a d c m^{- 1} .

The bracket coefficient is not itself the wave number in radians per centimetre. It represents spatial phase change in cycles per centimetre, so multiply by the full-turn factor:

0.0080 c y c l e s c m^{- 1} \times 2 π r a d c y c l e^{- 1} = 0.016 π r a d c m^{- 1} .

The amplitude, 2.0 cm, sets the maximum displacement. It does not enter the phase-difference calculation because it multiplies the cosine rather than appearing inside its argument.

How do you subtract the phases and calculate option B?

At the same instant, the time term and initial phase are identical at both points and cancel on subtraction. Only the spatial term remains. Calculate the positive magnitude requested by the options, using the separation in centimetres.

Take the second point at a larger coordinate and compare both phases at the same time: $x_{2} - x_{1} = Δ x > 0 .$

ϕ_{2} - ϕ_{1} & = [20 π t - k x_{2} + 0.70 π] & - [20 π t - k x_{1} + 0.70 π] & = - k (x_{2} - x_{1}) .

The negative sign means that phase decreases as position increases at that instant. The options ask for the phase-difference magnitude, so use:

Δ ϕ = | ϕ_{2} - ϕ_{1} | = k Δ x .

Convert the separation before substitution:

Δ x = 0.5 m \times 100 c m m^{- 1} = 50 c m .

Now multiply by the wave number:

Δ ϕ & = 2 π \times 8 \times 10^{- 3} \times 50 & = 2 π \times 0.4 & = 0.8 π r a d .

Final answer: B)

Δ ϕ = 0.8 π r a d

Check that the length units cancel, leaving radians:

(r a d c m^{- 1}) (c m) = r a d .

Read the wave number directly rather than calculating wavelength first: it is shorter and follows the official method. For the bank’s 60-second target, use this sequence: identify the wave number, convert the separation, multiply.

Why does using 0.5 directly produce option D?

Using 0.5 without conversion mixes a distance in metres with a wave number in radians per centimetre. The arithmetic reproduces option D, but the units do not support the substitution:

2 π \times 8 \times 10^{- 3} \times 0.5 = 0.008 π .

Keeping the centimetre-based wave number makes this substitution treat the separation as 0.5 cm, rather than the stated 0.5 m. The result is one-hundredth of the correct magnitude:

\frac{0.008 π}{0.8 π} = \frac{1}{100} .

Repair rule: the numerical distance must use the length unit attached to the spatial coefficient. Write that unit beside the wave number before inserting any distance.

As a check after the official calculation, consistent SI units give the same answer:

k = 1.6 π r a d m^{- 1}, k (0.5 m) = 0.8 π r a d .

How do spatial separation and elapsed time change the phase?

Different positions at one instant require the wave number multiplied by distance. Different times at one position require angular frequency multiplied by elapsed time. The following are original practice variations based on the supplied wave, not additional verified NEET PYQs.

What is the phase difference between points 25 cm apart at one instant?

The phase-difference magnitude for practice question 1, using the same supplied wave, is:

0.4 π r a d

The time and initial-phase terms cancel. The separation already matches the wave number’s centimetre unit:

Δ ϕ & = k Δ x & = 2 π \times 0.0080 \times 25 & = 0.4 π r a d .

What is the phase change over 0.025 seconds at one fixed position?

For practice question 2, the same wave’s phase increases over 0.025 s by:

0.5 π r a d

At a fixed position, the spatial term cancels. Read the coefficient of time from the complete phase:

ω = 2 π \times 10 = 20 π r a d s^{- 1} .

Δ ϕ = ω Δ t = 20 π \times 0.025 = 0.5 π r a d .

Choose the coefficient according to what changes:

Same time, different positions: use the spatial phase-difference magnitude. $Δ ϕ = k Δ x$
Same position, later time: use the temporal phase change. $Δ ϕ = ω Δ t$

For any two events in this wave, the general subtraction rule is:

Δ ϕ_{s i g n e d} = ω Δ t - k Δ x .

The first practice question sets the time change to zero. The second sets the position change to zero. Identify what stays fixed before choosing the coefficient.

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Frequently asked questions

How do you calculate phase difference between two points at the same time?

For a travelling harmonic wave, the phase-difference magnitude at the same instant is Δφ = k|Δx|, where k is the wave-number magnitude. The time term and initial phase cancel when you subtract the phases. In this question, k = 0.016π rad/cm and the separation is 50 cm, giving 0.8π rad, option B.

How do you find the wave number from the given wave equation?

For y = 2.0 cos[2π(10t − 0.0080x + 0.35)], expand the entire cosine argument before reading the coefficient of position. The wave-number magnitude is 2π × 0.0080 = 0.016π rad/cm because x is measured in centimetres. The coefficient 0.0080 alone is in cycles per centimetre, not radians per centimetre.

Why is option D wrong in the phase-difference question?

Option D results from substituting 0.5 directly into a calculation whose wave number is in rad/cm. The stated separation is 0.5 m, so it must first become 50 cm. Skipping this conversion gives 0.008π instead of 0.8π rad, making the answer 100 times too small.

When should I use kΔx instead of ωΔt for phase difference?

Use k|Δx| for the phase-difference magnitude between positions at the same instant. Use ωΔt for the phase change over time at one fixed position in the supplied wave. If both position and time change, use the signed relation Δφ = ωΔt − kΔx.