To solve this problem, we need to understand the concept of resonance in air columns that are closed at one end. In a closed pipe, the fundamental frequency of vibration is determined by the formula:
f = \frac{v}{4L}
where f is the frequency, v is the velocity of sound, and L is the length of the air column.
Let's denote L_1 = 100 \text{ cm} = 1 \text{ m} and L_2 = 120 \text{ cm} = 1.2 \text{ m}. The frequencies of these two columns in their fundamental modes are:
f_1 = \frac{v}{4 \times 1} = \frac{v}{4}
f_2 = \frac{v}{4 \times 1.2} = \frac{v}{4.8}
The problem states that these two frequencies produce 15 beats per second:
|f_1 - f_2| = 15
Substitute the expressions for f_1 and f_2:
\left|\frac{v}{4} - \frac{v}{4.8}\right| = 15
Calculate the difference inside the absolute value:
\frac{v}{4} - \frac{v}{4.8} = \frac{v \cdot 4.8 - v \cdot 4}{4 \cdot 4.8}
\frac{v(4.8 - 4)}{19.2} = \frac{0.8v}{19.2} = \frac{v}{24}
Therefore, the equation becomes:
\frac{v}{24} = 15 \quad \Rightarrow \quad v = 15 \times 24 = 360 \text{ m/s}
Thus, the velocity of sound in the air column is 360 m/s.
The correct answer is 360 \text{ m/s}.
Similar Questions
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Identify the correct match below :
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Important Questions from Oscillations and Waves
- Two tuning forks $A$ and $B$ are sounded together giving rise to $8$ beats in $2\text{ s}$. When fork $A$ is loaded with wax, the beat frequency is reduced to $4$ beats in $2\text{ s}$. If the original frequency of tuning fork $B$ is $380\text{ Hz}$ then original frequency of tuning fork $A$ is _________ $\text{Hz}$.
- 5 beats/second are heard when a turning fork is sounded with a sonometer wire under tension, when the length of the sonometer wire is either 0.95 m or 1 m. The frequency of the fork will be
- A sinusoidal wave of wavelength 7.5 cm travels a distance of 1.2 cm along the x-direction in 0.3 sec. The crest P is at x=0 at t=0 sec and maximum displacement of the wave is 2 cm. Which equation correctly represents this wave ?
A particle is subjected to two simple harmonic motions as :
$x_1 = \sqrt{7} \sin 5t \text{ cm}$ and $x_2 = 2\sqrt{7} \sin \left(5t + \frac{\pi}{3}\right) \text{ cm}$
where x is displacement and t is time in seconds. The maximum acceleration of the particle is $x \times 10^{-2} \text{ ms}^{-2}$. The value of x is :- The equation of a transverse wave is $y = y_0 \sin 2\pi(ft - \frac{x}{\lambda})$. If the maximum particle velocity be four times that of wave velocity then
