Question 333: Waves

Question

A set of 28 turning forks is arranged in an increasing order of frequencies. Each fork produces ' $x$ ' beats per second with the preceding fork and the last fork is an octave of the first. If the frequency of the $12^{	ext {th }}$ fork is 152 Hz , the value of ' $x$ ' (no. of beats per second) is

Accepted Answer

A set of 28 tuning forks is arranged by increasing frequencies, each producing ' $x$ ' beats per second with the preceding fork. The frequency of the last fork is double that of the first, meaning it's an octave higher.
Given the frequency of the 12
th
fork is 152 Hz, our task is to find the value of ' $x$ '.
Let's denote the frequencies as $	ext{n}_1$ through $	ext{n}_{28}$. Each tuning fork has $x$ beats with the previous one.
For the 12
th
fork, we have:
$ 	ext{n}_{12} = 	ext{n}_1 + 11x $
Since $	ext{n}_{12}$ is 152 Hz, we can write:
$ 152 = 	ext{n}_1 + 11x \quad 	ext{(equation 1)} $
For the last, or 28
th
fork:
$ 	ext{n}_{28} = 	ext{n}_1 + 27x $
The condition that the 28
th
fork is an octave higher means:
$ 	ext{n}_{28} = 2	ext{n}_1 $
Thus, we have:
$ 2	ext{n}_1 = 	ext{n}_1 + 27x \quad \Rightarrow \quad 	ext{n}_1 = 27x $
Substitute this expression for $	ext{n}_1$ into equation (1):
$ 152 = 27x + 11x $
Simplify to find $x$:
$ 152 = 38x \quad \Rightarrow \quad x = \frac{152}{38} = 4 $
Hence, the number of beats per second, ' $x$ ', is 4.

Answer

2

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4

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6

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8

Waves

A set of 28 turning forks is arranged in an increasing order of frequencies. Each fork produces ' x ' beats per second with the preceding fork and the last fork is an octave of the first. If the frequency of the 12^{\text {th }} fork is 152 Hz , the value of ' x ' (no. of beats per second) is

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