The differential equation of a wave is
a.$\dfrac{{{d^2}y}}{{d{t^2}}} = {v^2}\dfrac{{{d^2}y}}{{d{x^2}}}$
b.$\dfrac{{{d^2}y}}{{d{x^2}}} = {v^2}\dfrac{{{d^2}y}}{{d{t^2}}}$
c.$\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{1}{v}\dfrac{{{d^2}y}}{{d{t^2}}}$
d.$\dfrac{{{d^2}y}}{{d{x^2}}} = - v\dfrac{{{d^2}y}}{{d{t^2}}}$
Answer
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Hint: The wave equation is a second-order linear partial differential equation for the portrayal of the waves as they occur in nature. There are different examples of wave equations such as electromagnetic wave equations, acoustic waves, etc.
Complete answer:
The progressive wave equation is
$y = A\sin (\omega t - kx)$ ……………… (1)
From equation (1) $y = A\sin (\omega t - kx)$
Now double differentiate this equation with respect to $t$ :
$\dfrac{{{d^2}y}}{{d{t^2}}} = \dfrac{d}{{dx}}(A( - \cos (\omega t - kx)(w))$
$\dfrac{{{d^2}y}}{{d{t^2}}} = - \dfrac{d}{{dx}}(Aw\cos (\omega t - kx))$
$\dfrac{{{d^2}y}}{{d{t^2}}} = - A{w^2}\sin (\omega t - kx)$ ………….. (2)
From equation (1) and (2)
We get $\dfrac{{{d^2}y}}{{d{t^2}}} = - {w^2}y$ …….. (3)
Now double differentiate this equation with respect to $x$ :
$\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}(A( - \cos (\omega t - kx)( - k))$
$\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}(Ak\cos (\omega t - kx))$
$\dfrac{{{d^2}y}}{{d{x^2}}} = - A{k^2}\sin (\omega t - kx)$ ……….. (4)
From equation (1) and (4)
We get,
$\dfrac{{{d^2}y}}{{d{x^2}}} = - {k^2}y$ …………. (5)
Comparing equations (4) and (5)
We get,
$\dfrac{1}{{{w^2}}}\dfrac{{{d^2}y}}{{d{t^2}}} = \dfrac{1}{{{k^2}}}\dfrac{{{d^2}y}}{{d{x^2}}}$
$\dfrac{{{d^2}y}}{{d{t^2}}} = \dfrac{{{w^2}}}{{{k^2}}}\dfrac{{{d^2}y}}{{d{x^2}}}$ ……………… (6)
From relation $v = \dfrac{\omega }{k}$
$\dfrac{{{d^2}y}}{{d{t^2}}} = {v^2}\dfrac{{{d^2}y}}{{d{x^2}}}$
So the correct option is (a).
Note:
Wave is the propagation of disturbance from place to place in a regular and ordered manner. The most familiar waves are light, sound, and the motion of subatomic particles.
There are two kinds of waves, the first one is transverse and the other are longitudinal waves. Transverse waves are observed in water where Sound is an example of longitudinal waves.
Waves can move immense distances even though oscillation at one point is very small. For example, we can hear a thunderclap from kilometers away.
Complete answer:
The progressive wave equation is
$y = A\sin (\omega t - kx)$ ……………… (1)
From equation (1) $y = A\sin (\omega t - kx)$
Now double differentiate this equation with respect to $t$ :
$\dfrac{{{d^2}y}}{{d{t^2}}} = \dfrac{d}{{dx}}(A( - \cos (\omega t - kx)(w))$
$\dfrac{{{d^2}y}}{{d{t^2}}} = - \dfrac{d}{{dx}}(Aw\cos (\omega t - kx))$
$\dfrac{{{d^2}y}}{{d{t^2}}} = - A{w^2}\sin (\omega t - kx)$ ………….. (2)
From equation (1) and (2)
We get $\dfrac{{{d^2}y}}{{d{t^2}}} = - {w^2}y$ …….. (3)
Now double differentiate this equation with respect to $x$ :
$\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}(A( - \cos (\omega t - kx)( - k))$
$\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}(Ak\cos (\omega t - kx))$
$\dfrac{{{d^2}y}}{{d{x^2}}} = - A{k^2}\sin (\omega t - kx)$ ……….. (4)
From equation (1) and (4)
We get,
$\dfrac{{{d^2}y}}{{d{x^2}}} = - {k^2}y$ …………. (5)
Comparing equations (4) and (5)
We get,
$\dfrac{1}{{{w^2}}}\dfrac{{{d^2}y}}{{d{t^2}}} = \dfrac{1}{{{k^2}}}\dfrac{{{d^2}y}}{{d{x^2}}}$
$\dfrac{{{d^2}y}}{{d{t^2}}} = \dfrac{{{w^2}}}{{{k^2}}}\dfrac{{{d^2}y}}{{d{x^2}}}$ ……………… (6)
From relation $v = \dfrac{\omega }{k}$
$\dfrac{{{d^2}y}}{{d{t^2}}} = {v^2}\dfrac{{{d^2}y}}{{d{x^2}}}$
So the correct option is (a).
Note:
Wave is the propagation of disturbance from place to place in a regular and ordered manner. The most familiar waves are light, sound, and the motion of subatomic particles.
There are two kinds of waves, the first one is transverse and the other are longitudinal waves. Transverse waves are observed in water where Sound is an example of longitudinal waves.
Waves can move immense distances even though oscillation at one point is very small. For example, we can hear a thunderclap from kilometers away.
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