Question

The pressure wave, $P=0.01Sin\left[ 1000t-3x \right]N{{m}^{-2}}$, corresponds to the second produced by the vibrating blade on a day when the atmospheric pressure is ${{0}^{\circ }}C$. On some other day when the temperature is T, the speed of sound produced by the same blade and at the same frequency is found to be $336m{{s}^{-1}}$. Approximate value of T is
$\begin{aligned} & a){{15}^{\circ }}C \\\ & b){{12}^{\circ }}C \\\ & c){{4}^{\circ }}C \\\ & d){{11}^{\circ }}C \\\ \end{aligned}$

Answer 1

In the above question it is given to us how the pressure wave varies with time ‘t’. We will first compare the above given wave equation to its general form and then obtain the value of constant of propagation and the angular frequency of the wave in order to determine the velocity of the velocity of sound at ${{0}^{\circ }}C$. The velocity of sound is directly proportional to the surrounding temperature. Hence we will accordingly take the ratio between the velocity at ${{0}^{\circ }}C$ and velocity at T to determine the temperature.

Formula used:
$P={{P}_{o}}Sin\left[ \omega t-kx \right]N{{m}^{-2}}$
$v=\dfrac{\omega }{k}$
$v\propto \sqrt{T}$

Complete step-by-step answer:
To begin with, let us first write the wave equation for pressure in its general form.
Let us say the pressure of the sound wave has a maximum value of ${{P}_{o}}$ . Let the angular velocity be $\omega$and the wave propagation constant be k. Therefore the pressure of the wave at time ‘t’ and position ‘x’ is given by,
$P={{P}_{o}}Sin\left[ \omega t-kx \right]N{{m}^{-2}}$
In the above question the wave equation is given to us as,$P=0.01Sin\left[ 1000t-3x \right]N{{m}^{-2}}$
Comparing the respective physical quantities in the above equation, their respective values we get as,
${{P}_{\circ }}=0.01\text{, }\omega \text{=1000rad/s and k=3rad/m}$. The speed of the sound (v)of the above wave in general is given by,
$v=\dfrac{\omega }{k}$
Hence the speed of the sound at ${{0}^{\circ }}C$ from the above wave equation we get as,
$\begin{aligned} & v=\dfrac{\omega }{k} \\\ & \Rightarrow v=\dfrac{1000rad/s}{3rad/m} \\\ & \Rightarrow v=\dfrac{1000}{3}m/s \\\ \end{aligned}$
The speed of the sound wave is directly proportional to the square root of the surrounding temperature(T). Mathematically this can be represented as,
$\begin{aligned} & v\propto \sqrt{T} \\\ & \Rightarrow v=a\sqrt{T} \\\ \end{aligned}$
Where a is constant of proportionality. At ${{0}^{\circ }}C$ i.e. is at 273K, the velocity of the sound we found as 1000/3 and it is given to us that the velocity of the sound produced by the same blade at temperature T is 336m/s. Hence from the above velocity equation, taking the ratio of the two cases we get T as,
$\begin{aligned} & \dfrac{\dfrac{1000}{3}}{336}=\dfrac{a\sqrt{273}}{a\sqrt{T}} \\\ & \Rightarrow \dfrac{1000}{3\times 336}=\dfrac{\sqrt{273}}{\sqrt{T}} \\\ & \Rightarrow \sqrt{T}=\dfrac{\sqrt{273}(3\times 336)}{1000} \\\ & \Rightarrow T=277.41K \\\ & \Rightarrow T={{4}^{\circ }}C \\\ \end{aligned}$
Hence the correct answer of the above question is option c.

So, the correct answer is “Option C”.

Note: The relation between the Kelvin scale and the Celsius scale is as follows. Let the reading on the Celsius scale be ${{T}^{\circ }}C$. Therefore the corresponding temperature reading on the Kelvin scale i.e. ${{M}^{\circ }}K$ is equal to ${{M}^{\circ }}K={{T}^{\circ }}C+273$ . The velocity of sound depends on the constant of proportionality ’a’ as well which is dependent on the medium and the frequency of the sound.