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Different Forms of Equations

A transverse ...

Question

A transverse wave is propagating along $+ x$ direction. At $t = 2 s$ , the particle at $x = 4$ $m$ is at $y = 2 m m$ . With the passage of time its $y$ coordinate increases and reaches to a maximum of $4 m m$ . The wave equation is (using $ω$ and $k$ with their usual meanings)

y = 4 sin (ω (t + 2) + k (x - 2) + \frac{π}{6}]

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y = 4 sin (ω (t + 2) + k x + \frac{π}{6}]

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y = 4 sin (ω (t - 2) + k (x - 4) + \frac{5 π}{6}]

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y = 4 sin (ω (t - 2) - k (x - 4) + \frac{π}{6}]

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Solution

The correct option is D $y = 4 sin (ω (t - 2) - k (x - 4) + \frac{π}{6}]$
We know the general equation as
$y = A s i n (ω t - k x + ϕ)$
So as given at $t = 2 s$
$y = 2 m m$
$x = 4 m m$
Amplitude is given as $4 m m$
SHM equation of the particle is
$\Rightarrow 2 = 4 s i n (2 ω - 4 k + ϕ)$
Solving this we get $ϕ = - 2 ω + 4 k + \frac{π}{6}$
Putting this value in equation in wave equation
$\Rightarrow y = 4 s i n (ω t - k x + (- 2 ω + 4 k + \frac{π}{6}$
$\Rightarrow y = 4 s i n (ω (t - 2) - k (x - 4) + \frac{π}{6}$

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Similar questions

+ x

t = 2 s

x = 4

m

y = 2 m m

y

4 m m

ω

k

-

x -

A, ω

k

x = \frac{λ}{4}

y (t) = A sin ω t

y_{1} = 8 sin (ω t + k x); y_{2} = 6 sin (ω t + \frac{π}{2} + k x); y_{3} = 4 sin (ω t + π + k x);

y_{4} = 2 sin (ω t + \frac{3 π}{2} + k x);

y_{5} = 4 \sqrt{2} sin (ω t - k x + \frac{π}{4})

Y = 2 sin (k x - 2 t)

y

y = f (x, t)

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