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Phase or path difference between a sine wave and a cosine wave

This type of question, asked 6 times: 2 May 2023 S1, 3 May 2023 S1, 13 Apr 2026 S2, 20 Apr 2026 S2, 12 May 2026 S1, 13 May 2026 S2. All 6 questions below, oldest first.

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  1. Question 1

    2 May 2023 S1

    15
    Two simple harmonic progressive waves have displacements y1⃗=a1sin⁡(2πxλ−ωt)\vec{y_1} = a_1 \sin\left(\frac{2\pi x}{\lambda} - \omega t\right) and y2=a2cos⁡(2πxλ−ωt+ϕ)y_2 = a_2 \cos\left(\frac{2\pi x}{\lambda} - \omega t + \phi\right). What is the phase difference between two waves?

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  2. Question 2

    3 May 2023 S1

    3
    The path difference between two waves Y1=a1sin⁡ ⁣(ωt−2πxλ)Y_1=a_1\sin\!\left(\omega t-\frac{2\pi x}{\lambda}\right) and Y2=a2cos⁡ ⁣(ωt−2πxλ+ϕ)Y_2=a_2\cos\!\left(\omega t-\frac{2\pi x}{\lambda}+\phi\right) is

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  3. Question 3

    13 Apr 2026 S2

    122
    Two waves are represented by the equations, y1=asin⁡(ωt+kx+0.57)y_1 = a\sin(\omega t + kx + 0.57) m and y2=acos⁡(ωt+kx)y_2 = a\cos(\omega t + kx) m, where xx is in metre and tt is in second. What is the phase difference between them? (π=3.14)(\pi = 3.14)

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  4. Question 4

    20 Apr 2026 S2

    123
    Two waves are represented as y1=a1sin⁡(ωt−2πxλ)y_1 = a_1\sin\left(\omega t - \dfrac{2\pi x}{\lambda}\right) and y2=a2cos⁡(ωt−2πxλ+π6)y_2 = a_2\cos\left(\omega t - \dfrac{2\pi x}{\lambda} + \dfrac{\pi}{6}\right) The path difference between two waves is

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  5. Question 5

    12 May 2026 S1

    122
    Two simple harmonic progressive waves have displacements y1=5sin⁡(2πxλ−ωt)y_1 = 5\sin\left(\dfrac{2\pi x}{\lambda} - \omega t\right) and y2=5cos⁡(2πxλ−ωt+π6)y_2 = 5\cos\left(\dfrac{2\pi x}{\lambda} - \omega t + \dfrac{\pi}{6}\right). The phase difference between the two waves is

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  6. Question 6

    13 May 2026 S2

    122
    The path difference between two waves Y1=a1sin⁡(ωt−2πxλ)Y_1 = a_1\sin\left(\omega t - \dfrac{2\pi x}{\lambda}\right) and Y2=a2cos⁡(ωt−2πxλ+Φ)Y_2 = a_2\cos\left(\omega t - \dfrac{2\pi x}{\lambda} + \Phi\right) is

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