Two particles of a medium disturbed by the wave propagation are at x 1-0 and x 2-1 cm- The displacement of the particles is given by the equation -y 1-2 sin 3 - t cm - y 2-2 sin 3 - t - - 8 cm t is in seconds The displacement equation of the particles at t -1 sec and x -4 cm is-A- 2 cmB- 1 cmC- 0-5 cmD- 0-25 cm

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Standard XII

Physics

Differential Equation of Wave

Two particles...

Question

Two particles of a medium disturbed by the wave propagation are at x1 = 0 and x2 = 1 cm. The displacement of the particles is given by the equation : $y_{1} = (2 s i n 3 π t)$ cm; $y_{2} = 2 s i n (3 π t - π / 8) c m$ (t is in seconds)
The displacement equation of the particles at t = 1 sec and x = 4 cm is...

2 cm

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1 cm

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0.5 cm

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0.25 cm

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Solution

The correct option is A 2 cm
As the motions of particles are simple harmonic, amplitude = 2cm , $ω$ (Angular frequency) =3 $π$ rad/sec. As the phase difference of particles separated by 1 cm is $π$ /8
$λ = \frac{8}{π} \times 2 π = 16$ cm
Hence, wave velocity $= \frac{ω}{2 π} \times λ = 24$ cm/s
The wave equation, then would be
$y = (2) s i n [\frac{2 π}{16} (24 t - x)]$ cm.
Putting t = 1 sec and x = 4 cm, we get y = 2 cm

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Two particles of a medium disturbed by the wave propagation are at x1 = 0 and x2 = 1 cm. The displacement of the particles is given by the equation : y1=(2 sin 3πt)cm; y2=2sin(3πt−π/8)cm (t is in seconds) The displacement equation of the particles at t = 1 sec and x = 4 cm is...