Question 231: Waves

Question

The distance between two consecutive points with phase difference of $45^{\circ}$ in a wave of frequency 300 Hz is 4.0 m . The velocity of the travelling wave is (in $\mathrm{km} / \mathrm{s}$ )

Accepted Answer

We know that the phase difference $\Delta \phi$ between two points separated by distance $\Delta x$ in a wave is given by:
$$ \Delta \phi = \frac{2\pi}{\lambda} \Delta x $$
We can rearrange this formula to solve for the wavelength $\lambda$:
$$ \lambda = \frac{2\pi}{\Delta \phi} \Delta x $$
We are told that the phase difference is $45^\circ$. To use the formula, we need to convert $45^\circ$ into radians:
$$ 45^\circ = \frac{45}{180} 	imes \pi = \frac{\pi}{4} $$
Now, plug this value and $\Delta x = 4\,\mathrm{m}$ into the wavelength formula:
$$ \lambda = \frac{2\pi}{\frac{\pi}{4}} 	imes 4 $$
Calculate the fraction first: $\frac{2\pi}{\frac{\pi}{4}} = 2\pi 	imes \frac{4}{\pi} = 8$.
So,

$$ \lambda = 8 	imes 4 = 32~\mathrm{m} $$
We know frequency $f = 300\,\mathrm{Hz}$ and wavelength $\lambda = 32\,\mathrm{m}$.
The velocity $v$ of the wave is given by:

$$ v = f \lambda = 300 	imes 32 = 9600\,\mathrm{m/s} $$
Change $v$ from meters per second to kilometers per second:

$$ 9600\,\mathrm{m/s} = 9.6\,\mathrm{km/s} $$

Answer

1.6

Answer

3.6

Answer

4.8

Answer

9.6

Waves

The distance between two consecutive points with phase difference of 45^{\circ} in a wave of frequency 300 Hz is 4.0 m . The velocity of the travelling wave is (in \mathrm{km} / \mathrm{s} )

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