Find an expression for the total average power radiated by a pulsating sphere of radius a. Discuss the power radiated as a function of \(\lambda / a\) for a constant amplitude of the velocity displacement. What semiquantitative conclusions can be made with regard to the frequency dependence of the radiation of sound from ordinary paper-cone loudspeakers?

To find the total average power radiated by a pulsating sphere of radius "a", we begin by considering the conservation of energy. For a pulsating sphere, the power emitted can be considered as outgoing energy flux per unit area, integrated over the sphere's surface. Mathematically, this can be written as: \(P = \oint P_r dS\) where P is the total power radiated, \(P_r\) is the power emitted per unit area, and dS is the area element. Next, we can derive the radiation pressure (force per unit area) exerted on the sphere's surface due to the pulsation. By considering the mechanical energy (kinetic and potential) associated with the pulsation, we can calculate the radiation pressure and then the power radiated: \(P_r = \frac{1}{2} \rho\omega^2 a^2 \xi^2\), where \(\rho\) is the fluid density, \(\omega\) is the angular frequency, and \(\xi\) is the radial displacement of the sphere's surface. To find the total power radiated, we integrate the power per unit area over the sphere's entire surface: \(P = \oint P_r dS = \frac{1}{2} \rho\omega^2 a^2 \xi^2 \oint dS\). Finally, we perform the integration by recognizing that dS is the surface area of the sphere: \(P = \frac{1}{2} \rho\omega^2 a^2 \xi^2 4\pi a^2\). This gives us the expression for the total power radiated by the pulsating sphere: \(P = 2\pi\rho\omega^2 a^4 \xi^2\).

Now, we need to examine the power radiated as a function of the ratio between wavelength and radius, \(\lambda/a\), for a constant amplitude of the velocity displacement. To do this, we rewrite the expression for the total power radiated using sound velocities and frequencies: \(P = 2\pi\rho\omega^2 a^4 \xi^2 = 2\pi\rho(\frac{2\pi f}{\lambda})^2 a^4 \xi^2\), where \(f\) is the frequency and \(\lambda\) is the wavelength. We can express the power radiated as a function of \(\lambda/a\) by solving for this ratio and plugging in the expression for P: \(\frac{\lambda}{a} = \frac{2\pi f}{\omega} \Rightarrow \lambda = \frac{2\pi f}{\omega}a\). Now, plug this value for \(\lambda\) into the expression for P: \(P = 2\pi\rho\omega^2 a^4 \xi^2 = 2\pi\rho(\frac{\omega}{2\pi f/a})^2 a^4 \xi^2\). We can simplify this expression and find that the total average power radiated by the sphere is proportional to theinverse square of \(\lambda/a\): \(P \propto \frac{1}{(\lambda/a)^2}\).

To understand the frequency dependence of the radiation of sound from ordinary paper-cone loudspeakers, we need to consider the expression we obtained for the total average power radiated: \(P \propto \frac{1}{(\lambda/a)^2}\). Since \(\lambda = c/f\) where c is the speed of sound, we can rewrite this expression as: \(P \propto f^2\). This shows that the total average power radiated by a pulsating sphere (such as a paper-cone loudspeaker) is proportional to the square of the frequency. This implies that at higher frequencies, loudspeakers will radiate more power, making them more efficient for reproducing high frequencies compared to low frequencies. In addition, the power radiated also depends on the square of the radius of the sphere, implying that larger loudspeakers can radiate more power and are thus more suitable for reproducing low-frequency sounds than smaller ones. We can conclude that the frequency dependence of sound radiation from ordinary paper-cone loudspeakers is mainly influenced by the frequency and the size of the loudspeaker.