Two waves, each possessing a frequency of 100 Hz , a wavelength of 2.0 cm , and an amplitude of 2.0 mm , travel in the same direction on a string. Determine the phase difference and the resultant wave amplitude for the

Correct answer: D. First scenario: 3π rad , 0 mm ; Second scenario: 4π rad , 4.0 mm Given f = 100 Hz , λ = 2.0 cm , A = 2.0 mm . For the first scenario, the time difference is Δ t = 0.015 s . The phase difference due to the time difference is Δ φ_1 = 2π f Δ t . Δ φ_1 = 2π × 100 × 0.015 = 3π rad The resultant amplitude is A_ R1 = A^2 + A^2 + 2A^2 (Δ φ_1) . A_ R1 = 2A^2 + 2A^2 (3π) = 2A^2 - 2A^2 = 0 mm For the second scenario, the path difference is Δ x = 4.0 cm . The phase difference due to the path difference is Δ φ_2 = (2π)/(λ) Δ x . Δ φ_2 = (2π)/(2.0) × 4.0 = 4π rad The resultant amplitude is A_ R2 = A^2 + A^2 + 2A^2 (Δ φ_2) . A_ R2 = 2A^2 + 2A^2 (4π) = 4A^2 = 2A = 4.0 mm

Concepts Of Physics MCQ Edition [Volume 1]

Exam: Concepts Of Physics MCQ Edition [Volume 1]
Subject: Physics
Chapter: Wave Motion and Waves on a String

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Wave Motion and Waves on a String

Two waves, each possessing a frequency of 100 Hz , a wavelength of...

Two waves, each possessing a frequency of

100 \text{ Hz}

, a wavelength of

2.0 \text{ cm}

, and an amplitude of

2.0 \text{ mm}

, travel in the same direction on a string. Determine the phase difference and the resultant wave amplitude for the following two scenarios:

First scenario: The second wave is produced

0.015 \text{ s}

later than the first wave at the identical location.
Second scenario: Both waves are produced simultaneously, while the first wave is generated at a distance of

4.0 \text{ cm}

behind the second wave.

Identify the correct values of (phase difference, resultant amplitude) for the first and second scenarios, respectively.

Options

AFirst scenario: $4\pi \text{ rad}$ , $4.0 \text{ mm}$ ; Second scenario: $3\pi \text{ rad}$ , $0 \text{ mm}$
BFirst scenario: $3\pi \text{ rad}$ , $4.0 \text{ mm}$ ; Second scenario: $4\pi \text{ rad}$ , $0 \text{ mm}$
CFirst scenario: $1.5\pi \text{ rad}$ , $0 \text{ mm}$ ; Second scenario: $2\pi \text{ rad}$ , $4.0 \text{ mm}$
DFirst scenario: $3\pi \text{ rad}$ , $0 \text{ mm}$ ; Second scenario: $4\pi \text{ rad}$ , $4.0 \text{ mm}$

Correct answer

D. First scenario: $3\pi \text{ rad}$ , $0 \text{ mm}$ ; Second scenario: $4\pi \text{ rad}$ , $4.0 \text{ mm}$

Step-by-step solution

Given

f = 100 \text{ Hz}

\lambda = 2.0 \text{ cm}

A = 2.0 \text{ mm}

.

For the first scenario, the time difference is

\Delta t = 0.015 \text{ s}

.
The phase difference due to the time difference is

\Delta \phi_1 = 2\pi f \Delta t

\Delta \phi_1 = 2\pi \times 100 \times 0.015 = 3\pi \text{ rad}

The resultant amplitude is

A_{R1} = \sqrt{A^2 + A^2 + 2A^2 \cos(\Delta \phi_1)}

A_{R1} = \sqrt{2A^2 + 2A^2 \cos(3\pi)} = \sqrt{2A^2 - 2A^2} = 0 \text{ mm}

For the second scenario, the path difference is

\Delta x = 4.0 \text{ cm}

.
The phase difference due to the path difference is

\Delta \phi_2 = \dfrac{2\pi}{\lambda} \Delta x

\Delta \phi_2 = \dfrac{2\pi}{2.0} \times 4.0 = 4\pi \text{ rad}

The resultant amplitude is

A_{R2} = \sqrt{A^2 + A^2 + 2A^2 \cos(\Delta \phi_2)}

A_{R2} = \sqrt{2A^2 + 2A^2 \cos(4\pi)} = \sqrt{4A^2} = 2A = 4.0 \text{ mm}

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