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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

A standing wave produced on a string fixed at both ends has the...

A standing wave produced on a string fixed at both ends has the equation y=(0.4 cm)sin⁡[(0.314 cm−1)x]cos⁡[(600π s−1)t]y = (0.4 \text{ cm}) \sin[(0.314 \text{ cm}^{-1}) x] \cos[(600\pi \text{ s}^{-1})t]. Determine the smallest possible length of the string.

Options

  1. A40 cm40 \text{ cm}
  2. B10 cm10 \text{ cm}
  3. C5 cm5 \text{ cm}
  4. D20 cm20 \text{ cm}

Correct answer

B. 10 cm10 \text{ cm}

Step-by-step solution

The equation of the standing wave is given by y=Asin⁡(kx)cos⁡(ωt)y = A \sin(kx) \cos(\omega t).

Comparing the given equation with the standard form, we get k=0.314 cm−1k = 0.314 \text{ cm}^{-1}.

We know that k=2πλk = \dfrac{2\pi}{\lambda}.

2πλ=0.314⇒2×3.14λ=0.314⇒λ=20 cm\dfrac{2\pi}{\lambda} = 0.314 \Rightarrow \dfrac{2 \times 3.14}{\lambda} = 0.314 \Rightarrow \lambda = 20 \text{ cm}.

For a string fixed at both ends, the length of the string must be an integral multiple of λ2\dfrac{\lambda}{2}.

L=nλ2L = n\dfrac{\lambda}{2} where n=1,2,3,…n = 1, 2, 3, \ldots

For the smallest possible length, n=1n = 1.

L_\min = \dfrac{\lambda}{2} = \dfrac{20}{2} = 10 \text{ cm}.
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