Derive the expression for the speed of sound wave.
Answers
Answer:
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Explanation:
The derivation of the wave equation certainly varies depending on context. There is a particular simple physical setting for the derivation. Moreover, this setting is that of small oscillations on a piece of string in accordance with the Hooke’s law.
One must consider the forces acting on a small element of particular mass dm contained in a small interval dx. In case of a small displacement, the horizontal force is approximately zero. The vertical force happens to be
\Sigma F_{y}ΣF
y
= {T}’ sin\Theta_{2}T’sinΘ
2
– T sin\Theta_{1}sinΘ
1
= (dm)a = \mu dx \frac{\delta ^2 y}{\delta t^2}μdx
δt
2
δ
2
y
Here, µ refers to the mass density, \mu = \frac{\delta m}{\delta x}μ=
δx
δm
On the other hand, the horizontal force is approximately zero when displacements are small, T cos \theta^{1} \approx T{}’cos \theta_{2} \approx TTcosθ
1
≈T’cosθ
2
≈T. So,
-\frac{\mu dx\frac{\delta^{2}y}{\delta t^{2}}}{T}−
T
μdx
δt
2
δ
2
y
\approx≈ \frac{T{}’sin \theta_{2 + T sin \theta_{1}}}{T}
T
T’sinθ
2+Tsinθ
1
= \frac{{T}’sin\theta_{1}}{T}
T
T’sinθ
1
+ \frac{T sin\theta_{1}}{T}
T
Tsinθ
1
\approx \frac{{T}’ sin\theta_{2}}{T{}’ cos}≈
T’cos
T’sinθ
2
+ \frac{T sin\theta_{1}}{T cos\theta_{1}}
Tcosθ
1
Tsinθ
1
= tan\theta_{1} + tan\theta_{2}tanθ
1
+tanθ
2
But, tan \theta_{1}tanθ
1
+ tan \theta_{2}tanθ
2
= -\Delta \frac{\delta y}{\delta x}−Δ
δx
δy
, the difference happens to be between x and x + dx. This is because, the tangent is certainly equal to the slope in a geometrical manner. So, if one divides over dx, one finds
-\frac{\mu \frac{\delta^{2}y}{\delta t^{2}}}{T}−
T
μ
δt
2
δ
2
y
= \frac{tan \theta_{1} + tan \theta_{2}}{dx}
dx
tanθ
1
+tanθ
2
= -\frac{\Delta \frac{\delta y}{\delta x}}{dx}−
dx
Δ
δx
δy
The rightmost term above refers to the definition of the derivative with respect to x. This is because, the difference is over an interval dx, and therefore one has
\frac{\mu }{T} \frac{\delta ^{2}y}{\delta t^{2}}
T
μ
δt
2
δ
2
y
= \frac{\delta ^{2}y}{\delta x^{2}}
δx
2
δ
2
y
which happens to be exactly the wave equation in one dimension for velocity v = \sqrt{\frac{T}{\mu }}
μ
T
.
.
.
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