Question

A siren's $60-$hole disc rotates at a steady speed of $360$revolutions per minute. The sound that is emitted is in tune with a frequency tuning fork.

• A. $10\mathrm{Hz}$
• B. $360\mathrm{Hz}$
• C. $216\mathrm{Hz}$
• D. $60\mathrm{Hz}$

Answer 1

The correct option is B

$360\mathrm{Hz}$

Step 1: Given data:

The number of holes in the disk$=60$

Speed $=360\mathrm{rpm}$

Step 2: Formula used:

Frequency is the number of waves pass a fixed point per unit time.

Frequency of sound $f=\frac{\mathrm{Number}\mathrm{of}\mathrm{waves}}{\mathrm{second}}\dots \dots \dots \dots \dots \dots \dots \dots \left(1\right)$

Step 3: Calculation of frequency:

Convert revolutions per minute to rev per sec

Speed in rev per second is defined as

$\begin{array}{rcl}S& =& \frac{360}{60}\mathrm{revolutions}\mathrm{per}\mathrm{second}\\ & =& 6\mathrm{rps}\end{array}$

The number of waves produced in each revolution is determined by the number of holes.

The number of waves = $60$

Thus frequency can be calculated by multiplying the no of waves per revolution and speed in revolution per second that is,

$\begin{array}{rcl}Frequency& =& \frac{\mathrm{No}.\mathrm{of}\mathrm{waves}}{\mathrm{rev}}\times \frac{\mathrm{rev}}{\sec }\\ & =& \mathrm{No}.\mathrm{of}\mathrm{waves}\mathrm{per}\sec \end{array}$

From the formula of frequency from equation (1)

$\begin{array}{rcl}F\mathrm{requency}& =& \mathrm{No}.\mathrm{of}\mathrm{waves}\mathrm{per}\sec \\ & =& 60\times 6\\ & =& 360\mathrm{Hz}\end{array}$

Hence, option B is the correct answer.