9) A sinusoidal wave of frequency 500 Hz has a velocity of 350 m/sec. a) How far apart are (2) points that differ in phase by t/3 radians? b) What is the wavelength of the wave? c) What is the time interval between these (2) points? d) What is the period of the wave? e) What is the angular frequency? f) What is the constant of propagation?
9) A sinusoidal wave of frequency 500 Hz has a velocity of 350 m/sec. a) How far apart are (2) points that differ in phase by t/3 radians? b) What is the wavelength of the wave? c) What is the time interval between these (2) points? d) What is the period of the wave? e) What is the angular frequency? f) What is the constant of propagation?
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Transcribed Image Text:**Question 9: Analysis of a Sinusoidal Wave**
Consider a sinusoidal wave with a frequency of 500 Hz and a velocity of 350 m/s.
a) How far apart are two points that differ in phase by \(\pi/3\) radians?
b) What is the wavelength of the wave?
c) What is the time interval between these two points?
d) What is the period of the wave?
e) What is the angular frequency?
f) What is the constant of propagation?
**Detailed Explanation:**
- **Wavelength and Phase Difference:**
To find how far apart two points are that differ in phase by \(\pi/3\) radians, use the relationship between wavelength \(\lambda\), wave number \(k\), and phase difference.
- **Wavelength (\(\lambda\)):**
It is given by \(\lambda = \frac{v}{f}\), where \(v\) is the velocity and \(f\) is the frequency.
- **Time Interval:**
The time interval between these points can be determined through the relation of phase difference to time.
- **Period (\(T\)):**
The period of the wave is the reciprocal of the frequency.
- **Angular Frequency (\(\omega\)):**
Angular frequency is calculated by \(\omega = 2\pi f\).
- **Constant of Propagation (Wave Number \(k\)):**
It is determined by \(k = \frac{2\pi}{\lambda}\).
These concepts are crucial for understanding wave properties and behaviors in physics.
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