Question

A vibrating tuning fork tied to the end of a string 1.988 m long is whirled round in a circle. If it makes two revolutions in a second, calculate the ratio of the frequencies of the highest and the lowest notes heard by a stationary observer situated in the plane of rotation of tuning fork at a large distance from the tuning fork. Speed of sound is $350m{s}^{-1}$. (In both case, the observer is assumed to be along the line of velocity sound).

• A. 1.732
• B. 1.154
• C. 1.278
• D. 1

Answer 1

The correct option is B 1.154

Number of revolutions per second = 2
Radius of the circle = 1.988 m
Linear velocity of the tuning fork is
$v=\left(2\pi R\right)\times 2=4\times \frac{22}{7}\times 1.988=25m{s}^{-1}$
Let the listener is located on the left side at a large distance in the diagram.
$\therefore$ Apparent frequency, when the tuning fork is approaching the listener is
$f_1=\left(\frac{V}{V-V_S}\right)f_0=1.077f_0$
(2) Apparent frequency, when the tuning fork is moving away from the listener is $f_2=\left(\frac{v}{v+v_s}\right)f_0$
$f_2=\frac{350}{350+25}{f}_0=0.933f_0$
$f_1$ is the highest note and $f_2$ is the lowest note.
Now, $\frac{f_1}{f_2}=\frac{1.077}{0.933}=1.154$