To find the velocity of sound using the given problem, we'll utilize the principles of sound waves in organ pipes. Since both pipes are in fundamental modes, we'll use the fundamental frequency formulas for open and closed pipes.
The calculated velocity of sound \(v = 294 \, \text{m/s}\) is within the specified range (294, 294).
For a closed pipe of length \(L = 150 \, \text{cm} = 1.5 \, \text{m}\):
The fundamental frequency is given by:
\(f_c = \frac{v}{4L}.\)
For an open pipe of length \(L = 350 \, \text{cm} = 3.5 \, \text{m}\):
The fundamental frequency is given by:
\(f_o = \frac{v}{2L}.\)
Given that the beat frequency is:
\(|f_c - f_o| = 7 \, \text{Hz}.\)
Substituting:
\(\left| \frac{v}{4 \cdot 1.5} - \frac{v}{2 \cdot 3.5} \right| = 7.\)
Simplifying:
\(\left| \frac{v}{6} - \frac{v}{7} \right| = 7.\)
Solving for \(v\):
\(\frac{v}{42} = 7 \implies v = 42 \cdot 7 = 294 \, \text{m/s}.\)
The Correct answer is: 294
Let \( C_{t-1} = 28, C_t = 56 \) and \( C_{t+1} = 70 \). Let \( A(4 \cos t, 4 \sin t), B(2 \sin t, -2 \cos t) \text{ and } C(3r - n_1, r^2 - n - 1) \) be the vertices of a triangle ABC, where \( t \) is a parameter. If \( (3x - 1)^2 + (3y)^2 = \alpha \) is the locus of the centroid of triangle ABC, then \( \alpha \) equals: