Two vibrating tuning forks produce progressiv | vedclass

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(B) The angular frequencies are $\omega_1 = 500\pi \, 	ext{rad/s}$ and $\omega_2 = 506\pi \, 	ext{rad/s}$.Frequency $f = \frac{\omega}{2\pi}$,so $f_

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$6$

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(B) The angular frequencies are $\omega_1 = 500\pi \, 	ext{rad/s}$ and $\omega_2 = 506\pi \, 	ext{rad/s}$.Frequency $f = \frac{\omega}{2\pi}$,so $f_1 = \frac{500\pi}{2\pi} = 250 \, 	ext{Hz}$ and $f_2 = \frac{506\pi}{2\pi} = 253 \, 	ext{Hz}$.The number of beats per second is $\alpha = |f_2 - f_1| = |253 - 250| = 3 \, 	ext{beats/s}$.The amplitudes are $a_1 = 4$ and $a_2 = 2$.The intensity ratio between maxima and minima is given by $\beta = \frac{I_{\max}}{I_{\min}} = \left( \frac{a_1 + a_2}{a_1 - a_2} ight)^2$.Substituting the values,$\beta = \left( \frac{4 + 2}{4 - 2} ight)^2 = \left( \frac{6}{2} ight)^2 = 3^2 = 9$.Finally,$\beta - \alpha = 9 - 3 = 6$.

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