Question

The equation of a sound wave in air is given by

$p=\left(0.01N{m}^{-2}\right)sin\left[\right(1000s^{-1})t-(3.0m^{-1}\left)x\right]$

(a) Find the frequency, wavelength and the speed of sound wave in air.

(b) If the equilibrium pressure of air is $1.0\times {10}^{5}N{m}^{-2}$, what are the maximum and minimum pressures at a point as the wave passes through that point.

$$\begin{array}{cc}& \\ \left(i\right)2.1& \left(p\right)\text{frequency in Hz}\\ \left(ii\right)1.0\times {10}^5+0.01& \left(q\right)\text{wavelength in m}\\ \left(iii\right)333& \left(r\right)\text{speed in m/s}\\ \left(iv\right)160& \left(s\right)\text{Max p in}N{m}^{-2}\end{array}$$

• A. (p) - (iv); (q) - (i); (r) - (iii); (s) - (ii); (t) - (v)
• B. (p) - (i); (q) - (ii); (r) - (iii); (s) - (iv); (t) - (v)
• C. (p) - (iv); (q) - (v); (r) - (iii); (s) - (ii); (t) - (i)
• D. (p) - (v); (q) - (ii); (r) - (i); (s) - (iii); (t) - (iv)

Answer 1

The correct option is A

(p) - (iv); (q) - (i); (r) - (iii); (s) - (ii); (t) - (v)

Comparing with the standard form of a travelling wave

$P=P_{0}sin\omega (t-\frac{x}{v})$

(p) frequency =$\frac{\omega }{2\pi }$

$\omega =1000s^{-1}$

f = 160 Hz

(q) wavelength =$\frac{2\pi }{k}$

k =$\frac{\omega }{v}$ = 3 from given equation

$\lambda$ = = 2.1 m

(r) speed = $\lambda$f

= 2.1 $\times$ 160

= 333 m/s

(s) pressure amplitude is 0.01 N m${}^{-2}$

So maximum amplitude will be

$P_0+\Delta P_0=(1.0\times {10}^5+0.01)$

(t) Minimum amplitude

$P_0-\Delta P_0=(1.0\times {10}^5-0.01)$