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
A sub-atomic particle of mass \( 10^{-30} \) kg is moving with a velocity of \( 2.21 \times 10^6 \) m/s. Under the matter wave consideration, the particle will behave closely like (h = \( 6.63 \times 10^{-34} \) J.s)