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vodka [1.7K]
3 years ago
15

What is the difference between transverse and longitudinal waves? Transverse waves always travel at smaller speeds than do longi

tudinal waves. In a transverse wave the disturbance always occurs perpendicular to the direction in which the wave travels, whereas in a longitudinal wave the disturbance always occurs parallel to the line of travel of the wave. Transverse waves always carry more energy than do longitudinal waves. In a transverse wave the disturbance always occurs parallel to the line of travel of the wave, whereas in a longitudinal wave the disturbance always occurs perpendicular to the direction in which the wave travels. Transverse waves always have greater frequencies than do longitudinal waves.
Physics
1 answer:
gregori [183]3 years ago
8 0

in transverse wave the disturbance always occur perpendicularto the direction. in wich the wave travels. where as longitudinal wava is disturbance always occur parallel to the direction in wich the wave travels.

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To find the amplitude of the sound, we must first determine the wavelength and the phase difference between the two speakers.

For the wavelength;

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Thus; for both speakers; the wavelength of the sound is:

\dfrac{\lambda}{2} = (10+30) cm

\dfrac{\lambda}{2} = (40) cm

λ = 80 cm

The relation between the path difference(Δx) and the phase difference(Δ∅) is:

\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o

where;

Δx = 10 cm

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Δ∅ = π rad

∴

\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o

\pi \ rad  = \dfrac{2 \pi}{80 \ cm}(10 \ cm) + \Delta \phi_o

\pi \ rad  = \dfrac{2 \pi}{8}+ \Delta \phi_o

\pi \ rad  = \dfrac{ \pi}{4}+ \Delta \phi_o

\Delta \phi_o  =  \pi -\dfrac{ \pi}{4}

\Delta \phi_o  = \dfrac{ 4\pi - \pi}{4}

\Delta \phi_o  = \dfrac{ 3\pi}{4} \ rad

Suppose both speakers are placed side-by-side, then the path difference between the two speakers is: Δx = 0 cm

Thus, we have:

\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o

\Delta \phi = \dfrac{2 \pi}{\lambda}(0 \ cm ) + \dfrac{3 \pi}{4} \ rad

\Delta \phi = \dfrac{3 \pi}{4} \ rad

∴

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A = 2a \ cos \bigg (\dfrac{\Delta \phi }{2} \bigg)

A = 2a \ cos \bigg (\dfrac{\dfrac{3 \pi}{4} }{2} \bigg)

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A = 0.765a

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