Sound Wave Travel Time Calculation
Given Information
- Frequency (\(f\)): 4 kHz = \(4 \times 10^3\) Hz
- Wavelength (\(\lambda\)): 25 cm = 0.25 m
- Distance (\(d\)): 2.5 km = 2500 m
Calculations
First, calculate the speed of the sound wave (\(v\)) using the formula:
\(v = f \times \lambda\)
Substitute the given values:
\(v = (4 \times 10^3 \text{ Hz}) \times (0.25 \text{ m})\)
\(v = 1000 \text{ m/s}\)
Next, calculate the time (\(t\)) taken to travel the distance (\(d\)) using the formula:
\(t = \frac{d}{v}\)
Substitute the distance and calculated speed:
\(t = \frac{2500 \text{ m}}{1000 \text{ m/s}}\)
\(t = 2.5 \text{ s}\)
Result
The time taken for the sound wave to travel 2.5 km is 2.5 seconds.
Similar Questions
- In a longitudinal wave, the distance between two consecutive compressions and two consecutive rarefactions is called:
- If the frequency of a sound wave increases and the speed remains constant, its wavelength will:
- A broadcasting station transmits waves with a frequency of $71 \times 10^4\ Hz$ and a speed of $3 \times 10^8\ m/s$. The wavelength of the wave is:
Important Questions from Waves
Which of the following is related to Doppler effect?
- The velocity v(x) of a particle moving in one dimension is given by v(x) = v 0 sin \(\rm\left(\frac{\pi x}{x_0}\right) \) , where v 0 and x 0 are positive constants of appropriate dimensions. If the particle is initially at x/x 0 = ϵ, where |ϵ| ≪ 1, then, in the long time, it
The position of a particle in one dimension changes in discrete steps. With each step it moves to the right, however, the length of the step is drawn from a uniform distribution from the interval \(\left[ {{\rm{λ }}\,{\rm{ - }}\,\frac{{\rm{1}}}{{\rm{2}}}{\rm{w,}}\,{\rm{λ }}\,{\rm{ + }}\,\frac{{\rm{1}}}{{\rm{2}}}{\rm{w}}} \right] \) , where λ and w are positive constants. If X denotes the distance from the starting point after N steps, the standard deviation \(\sqrt {\left\langle {{X^2}} \right\rangle \, - {{\left\langle X \right\rangle }^2}} \) for large values of N is
A particle of mass m in one dimension is in the ground state of a simple harmonic oscillator described by a Hamiltonian \(\frac{{{{\rm{P}}^{\rm{2}}}}}{{{\rm{2m}}}}{\rm{ + }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{m}}{{\rm{\omega }}^{\rm{2}}}{{\rm{x}}^{\rm{2}}} \) in the standard notation. An impulsive force at time t = 0 suddenly imparts a momentum P0 = \(\sqrt {{\rm{hm\omega }}} \) to it. The probability that the particle remains in the original ground state is
In an elastic scattering process at an energy E, the phase shifts satisfy δ 0 ≈ 30°, δ 1≈ 10°, while the other phase shifts are zero. The polar angle at which the differential cross-section peaks is closest to
