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PhyMedAdv 2014 • Paper 1

Stationary Wave Modes on String Fixed at One End

One end of a taut string of length 3 m3\text{ m} extending along the xx-axis is held fixed at x=0x = 0. The propagation speed of waves in the string is 100 m s−1100\text{ m s}^{-1}. The other end vibrates along the yy-direction such that stationary waves are established along the string. Identify the possible waveform(s) of these stationary waves:

Options

A

y(t)=Asin⁡(πx6)cos⁡(50πt3)y(t) = A \sin\left(\frac{\pi x}{6}\right) \cos\left(\frac{50\pi t}{3}\right)

Correct
B

y(t)=Asin⁡(πx3)cos⁡(100πt3)y(t) = A \sin\left(\frac{\pi x}{3}\right) \cos\left(\frac{100\pi t}{3}\right)

C

y(t)=Asin⁡(5πx6)cos⁡(250πt3)y(t) = A \sin\left(\frac{5\pi x}{6}\right) \cos\left(\frac{250\pi t}{3}\right)

Correct
D

y(t)=Asin⁡(5πx2)cos⁡(250πt)y(t) = A \sin\left(\frac{5\pi x}{2}\right) \cos(250\pi t)

Correct

Topics & Concepts

WavesWave Equation

Step-by-Step Solution

To find the possible waveforms of the stationary waves set up in the string, we analyze the boundary conditions and the wave equation:

  1. Boundary Conditions:

    • At x=0x = 0, the string is fixed, which means there must be a displacement node at this point: y(0,t)=0y(0, t) = 0 This condition is satisfied by a spatial component of the form sin⁡(kx)\sin(kx).
    • At x=L=3 mx = L = 3\text{ m}, the end of the string is vibrating (driven along the yy-direction), so an antinode is established at this end for resonance: ∣sin⁡(kL)∣=1  ⟹  kL=(2n−1)π2for n=1,2,3,…\left|\sin(kL)\right| = 1 \implies kL = (2n - 1)\frac{\pi}{2} \quad \text{for } n = 1, 2, 3, \ldots

  2. Wavenumber and Frequency: Given L=3 mL = 3\text{ m}, the allowed values of the wavenumber kk are: k=(2n−1)π2L=(2n−1)π6 rad m−1k = \frac{(2n - 1)\pi}{2L} = \frac{(2n - 1)\pi}{6}\text{ rad m}^{-1}

    The wave speed is given as v=100 m s−1v = 100\text{ m s}^{-1}. Using the dispersion relation ω=vk\omega = v k: ω=100×(2n−1)π6=50(2n−1)π3 rad s−1\omega = 100 \times \frac{(2n - 1)\pi}{6} = \frac{50(2n - 1)\pi}{3}\text{ rad s}^{-1}

  3. Evaluating the Modes:

    • For n=1n = 1 (Fundamental Mode): k=π6 m−1,ω=50π3 rad s−1k = \frac{\pi}{6}\text{ m}^{-1}, \quad \omega = \frac{50\pi}{3}\text{ rad s}^{-1}   ⟹  y(x,t)=Asin⁡(πx6)cos⁡(50πt3)(Matches Option A)\implies y(x, t) = A \sin\left(\frac{\pi x}{6}\right) \cos\left(\frac{50\pi t}{3}\right) \quad \text{(Matches Option A)}

    • For n=3n = 3: k=5π6 m−1,ω=250π3 rad s−1k = \frac{5\pi}{6}\text{ m}^{-1}, \quad \omega = \frac{250\pi}{3}\text{ rad s}^{-1}   ⟹  y(x,t)=Asin⁡(5πx6)cos⁡(250πt3)(Matches Option C)\implies y(x, t) = A \sin\left(\frac{5\pi x}{6}\right) \cos\left(\frac{250\pi t}{3}\right) \quad \text{(Matches Option C)}

    • For n=8n = 8: k=15π6=5π2 m−1,ω=50(15)π3=250π rad s−1k = \frac{15\pi}{6} = \frac{5\pi}{2}\text{ m}^{-1}, \quad \omega = \frac{50(15)\pi}{3} = 250\pi\text{ rad s}^{-1}   ⟹  y(x,t)=Asin⁡(5πx2)cos⁡(250πt)(Matches Option D)\implies y(x, t) = A \sin\left(\frac{5\pi x}{2}\right) \cos(250\pi t) \quad \text{(Matches Option D)}

    • For Option B: Here k=π3 m−1k = \frac{\pi}{3}\text{ m}^{-1}, which gives a node at x=3 mx = 3\text{ m} since sin⁡(π×33)=sin⁡(π)=0\sin\left(\frac{\pi \times 3}{3}\right) = \sin(\pi) = 0. Since the end at x=3 mx = 3\text{ m} is vibrating, this does not represent the correct resonant stationary wave.

Therefore, the correct options are A, C, and D.

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