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OLEGan [10]
3 years ago
9

PLS HELP compare longitudinal and transverse waves

Physics
2 answers:
juin [17]3 years ago
7 0
If the vibrations of the particles of the medium conveying a wave are to and fro and parallel to the direction of propagation of the wave, then the wave is called a longitudinal wave.

Transverse wave: If the vibrations of the particles of the medium conveying a wave are perpendicular to the direction of propagation of the wave, the wave is called a transverse
babymother [125]3 years ago
6 0
<span>A transverse wave is one for which the direction of oscillation is perpendicular to the direction of propagation of the wave whereas, for longitudinalwaves oscillations are in the direction of propagation. Ripples in pond water move about the surface of water and they simultaneously move away from the point-0 too.</span><span>
Longitudinal waves include sound waves(vibrations in pressure, particle of displacement, and particle velocity propagated in an elastic medium) and seismic P-waves (created by earthquakes and explosions). In longitudinal waves, the displacement of the medium is parallel to the propagation of thewave.


</span>
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You have a spring that stretches 0.070 m when a 0.10-kg block is attached to and hangs from it at position y0. Imagine that you
olga nikolaevna [1]

Answer:

a) Δy = 0.144 m

b) W = 0.145 J

c) Us = 0.32 J

d) ymax = 0.144 m

Explanation:

a) First let's find the spring constant using Hooke's Law

F = k*Δy   ⇒  k = F/Δy

where

F = m*g = 0.1 kg*9.81 m/s² = 0.981 N

and  Δy = 0.07 m. Hence

k = 0.981 N/0.07 m = 14.014 N/m ≈ 14 N/m

In order to find the position of the block when we let it go, we need to find the force that caused this expansion in the spring, we know that the reading of the scale was 3 N and this reading includes the force we want to find and the weight of the block, therefore:

f = 3 N - F = 3 N - 0.981 N = 2.019 N

Now that we have found the force we can use Hooke's Law in order to find the position of the block

f = k*Δy   ⇒   Δy = f/k

⇒   Δy = 2.019 N/14 N/m

⇒   Δy = 0.144 m

b) First, notice that there are two kind of potential energy: the potential energy in the spring and the potential energy due to the gravitational field:

W = ΔU

W = ΔUs + ΔUg

W = (Usf - Usi) + (Ugf - Ugi)

Notice that

Us = 0.5*k*y²

where

yf = 0.07 m + 0.144 m = 0.214 m  and

yi = 0.07 m

and we will take the zero level to be the equilibrium position where the block was hanging at rest. Hence

W = 0.5*k*(yf² - yi²) + m*g*(0 - Δy)

⇒ W = 0.5*14 N/m*((0.214 m)² - (0.07 m)²) + (0.1 kg)*(9.81 m/s²)*(0 - 0.144 m)

⇒ W = 0.145 J

c) When we let the block go the spring was stretched by

y = 0.07 m + 0.144 m = 0.214 m

Therefore:

Us = 0.5*k*y²

⇒ Us = 0.5*14 N/m*(0.214 m)²

⇒ Us = 0.32 J

d) Because the position that we pulled the block to it is considered as the amplitude for the vibrational motion that will happen after we release the block, then the maximum height the particle will reach above the equilibrium position is

ymax = Δy = 0.144 m

 

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