Two sinusoidal waves are defined by the expressions, $y_{1} = (2.00) sin (20 x - 32 t)$ and $y_{2} = (2.00) sin (25 x - 40 t)$ where $y_{1}, y_{2}$ and $x$ are in centimeters and $t$ is in seconds. What is the positive value of $x$ , closest to the origin, for which the phase difference between the two waves is $π$ rad. at $t = 2 s$ ?

0.10 cm

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0.15 cm

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0.06 cm

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0.12 cm

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Solution

The correct option is C $0.06 cm$
The same values of $sin θ$ repeats after an interval of, integer multiple of $2 π$ rad

Therefore, the phase difference is equal to $π$ whenever, $Δ ϕ = π + 2 n π$ , for all integral values of $n$ .

Substituting this into the equation of phase difference, we get,
$π + 2 n π = Δ ϕ = ϕ_{1} - ϕ_{2}$
$π + 2 n π = (20 x - 32 t) - (25 x - 40 t)$
$π + 2 n π = - 5 x + 8 t$
At $t = 2 s$ ,
$π + 2 n π = - 5 x + 8 \times 2$
Or,
$5 x = 16 - π - 2 n π . . . . . . . . (i)$

The smallest positive value of $x$ will be only slightly greater than zero.

$\Rightarrow 16 - π - 2 n π \approx 0$

$2 n π = 16 - π$

$n = \frac{16 - π}{2 π}$

$∴ n \approx 2$

putting this value in $(i)$ , we get,

$5 x = 16 - π - 2 (2) π$

$\Rightarrow x = \frac{16 - 5 π}{5} = 0.059 cm$

Hence, $(A)$ is the correct answer.

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Two sinusoidal waves are defined by the expressions, y1=(2.00) sin (20x−32t) and y2=(2.00) sin (25x−40t) where y1,y2 and x are in centimeters and t is in seconds. What is the positive value of x, closest to the origin, for which the phase difference between the two waves is π rad. at t=2 s?