a tuning fork has a frequency of 300 hz and hence a period of 3.33×10⁻³ s. if the tuning fork is in an…

Explanation:

Step1: Recall the relativistic Doppler - effect formula

The formula for the relativistic Doppler effect when the source is moving away from the observer is $f = f_0\sqrt{\frac{1 - \beta}{1+\beta}}$, where $f_0$ is the proper frequency of the source, $\beta=\frac{v}{c}$, $v$ is the relative speed of the source with respect to the observer, and $c$ is the speed of light. Here, $f_0 = 300$ Hz and $\beta=0.620$.

Step2: Substitute the values into the formula

First, calculate the value inside the square - root: $\frac{1 - \beta}{1+\beta}=\frac{1 - 0.620}{1 + 0.620}=\frac{0.38}{1.62}\approx0.2346$. Then, $f = f_0\sqrt{\frac{1 - \beta}{1+\beta}}=300\times\sqrt{0.2346}$. $\sqrt{0.2346}\approx0.4844$, so $f = 300\times0.4844\approx145.32$ (This is wrong. We should use the formula for the source moving towards the observer $f = f_0\sqrt{\frac{1+\beta}{1 - \beta}}$ when the source is approaching in an optical - based measurement). Using the correct formula $f = f_0\sqrt{\frac{1+\beta}{1 - \beta}}$, where $\beta = 0.620$, $\frac{1+\beta}{1 - \beta}=\frac{1 + 0.620}{1-0.620}=\frac{1.62}{0.38}\approx4.2632$. Then $f=300\times\sqrt{4.2632}$. $\sqrt{4.2632}\approx2.0648$, and $f = 300\times2.0648 = 619.44$ (This is also wrong. Let's start over). The correct formula for the relativistic Doppler effect for a source moving at speed $v$ relative to an observer is $f=\frac{f_0}{\sqrt{1-\beta^{2}}}$ when the motion is non - radial. Here, $\beta = 0.620$, so $1-\beta^{2}=1-(0.620)^{2}=1 - 0.3844 = 0.6156$. Then $\sqrt{1-\beta^{2}}=\sqrt{0.6156}\approx0.7846$.

Step3: Calculate the observed frequency

$f=\frac{f_0}{\sqrt{1 - \beta^{2}}}=\frac{300}{0.7846}\approx382$ Hz.

Answer:

382 Hz