91Ó°ÊÓ

The displacement at t = 0 is zero, From the given graph, the wavelength is 0.20 m.

Formula to calculate the time period of oscillation is $\mathrm{T}=\frac{\mathrm{λ}}{\mathrm{v}}$Substitute 0.20 m for and 344 m/s for v to find T.

$\begin{array}{r}\mathrm{T}=\frac{0.20\mathrm{m}}{344\mathrm{m}/s}\\ =\left(5.81×{10}^{−4}\mathrm{s}\right)\left(\frac{{10}^3\mathrm{ms}}{1\mathrm{s}}\right)\\ =0.58\mathrm{ms}\end{array}$

The graph for the pressure as a function of t is shown below

Therefore, the graph for the pressure as a function of t is shown below

The formula to calculate pressure in a wave is $P\left(X\right)=−B\frac{dy\left(x\right)}{dx}$were, Bis the bulk modulus of the air. Then, the displacement is written as $y\left(x\right)=−\frac{1}{B}∫P\left(X\right)dx$

The equation of P (x) for the first half of the pressure versus x graph at t = 0 is given as,

$P\left(X\right)=mx+C$where,m is slope, C is the constant

From the figure, P (x)=40 at x=0 and P(x)=-40 at x =0.100m Then, C = 40

$\begin{array}{r}\mathrm{P}\left(\mathrm{X}\right)=\mathrm{mx}+\mathrm{C}\\ −40=\mathrm{m}\left(0.100\mathrm{m}\right)+40\\ \mathrm{m}=\frac{−80}{0.100\mathrm{m}}\\ =−800\end{array}$

Thus, the equation of the pressure is written as,

role="math" localid="1668253290745" $\begin{array}{r}P\left(X\right)=−800x+40\\ y\left(x\right)=−\left(\frac{1}{1.42×{10}^5\mathrm{Pa}}\right)∫(−800x+40)dx\\ =\left(\frac{1}{1.42×{10}^5\mathrm{Pa}}\right)\left(−800\frac{x^2}{2}+40x\right)\\ =2.81×{10}^{−5}{x}^2−2.81×{10}^{−4}x\end{array}$

The following graph shows the graph of displacement y in the sound wave as a function of x at t = with amplitude approximately 2.00 and —2.00.

Therefore, the graph of displacement y in the sound wave as a function of x is obtained as parabolic.

The maximum pressure of the air is 40 Pa, the bulk modulus of the air is 1.42 x 105 Pa and the speed of the sound in air is 344 m/s.

The following graph shows the graph of displacement y in the sound wave as a function of t at x = 0.

Therefore, the graph of displacement y as a function of t for x = 0 is obtained in figure Ill.

The maximum pressure of the air is 40 Pa, the bulk modulus of the air is $1.42×{10}^{2}Pa$and the speed of the sound in air is 344 m/s.

Formula to calculate the maximum velocity of the particle through which the sound wave is travelling is $V_{max}=\frac{P_{max}V}{B}$Substitute the values in the equation

$\begin{array}{r}{\mathrm{v}}_{max}=\frac{40\mathrm{Pa}×344\mathrm{m}/\mathrm{s}}{1.42×{10}^2\mathrm{Pa}}\\ =\left(0.0969\mathrm{m}×\frac{100\mathrm{cm}}{1\mathrm{m}}\right)\\ =9.69\mathrm{cm}/\mathrm{s}\end{array}$

Formula to calculate the maximum acceleration of the particle is $a_{max}=\frac{\left(\frac{dp}{dx}\right)}{p}$ Substitute the values we get,

$\begin{array}{r}{\mathrm{a}}_{max}=\frac{\left(\frac{80.0×\frac{\left(\frac{1\mathrm{kg}}{\mathrm{m}}\right)}{1\mathrm{Pa}}}{0.1\mathrm{m}}\right)}{1.20\mathrm{kg}/{\mathrm{m}}^3}\\ =667\mathrm{m}/{\mathrm{s}}^2\end{array}$

Thus, the maximum acceleration of the particle is$667\mathrm{m}/{\mathrm{s}}^2$

The maximum pressure of the air is 40 Pa, the bulk modulus of the air is 1.42 x 105 Pa and the speed of the sound in air is 344 m/s.

The generation of sound in loudspeaker is due to vibrating motion of the diaphragm of the speaker. The vibration of the diaphragm pushes or pulls its surrounding air and thus causes displacement. The displacement of the air changes with respect to air. Thus, the generation of sound in loudspeaker can be best describe by drawing a graph of displacement as a function of time, as shown in the solution of part

Therefore, the cone of the loudspeaker moves with the displacement as a function of time