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Alex Ar [27]
3 years ago
13

if two waves with equal amplitudes and wavelengths travel through a medium in such a way that a particular particle of the mediu

m is at the crest of one wave and at the trough of the other wave at the same time, what will happen to that particle?
Physics
2 answers:
Novosadov [1.4K]3 years ago
8 0
If two waves with equal amplitude and wavelength travel through a medium such that particular particle is at the crest of one wave and at the trough of another wave, then destructive interference will take place.Both waves will cancel each other out at that point such that a node will form in which the particle will remain at its rest position only and will not move. 
Irina-Kira [14]3 years ago
8 0

Answer:

As per the law of superposition we know that when two waves superimpose at a point on the medium then the resultant displacement of the medium particle is vector sum of the displacement due to each wave.

So here by superposition principle we can write

y_{net} = y_1 + y_2

now here we know that one of the wave reaches with crest while other wave reaches at trough position.

So we will have

y_1 = +A

y_2 = -A

so here by superposition we will have

y_{net} = A - A = 0

so after superposition the medium particle will come to its mean position.

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4.1 A steel spur pinion has a pitch of 5 teeth/in, 20 full-depth teeth, and a 20° pressure angle. The pinion runs at a speed of
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Answer:

15.07 ksi

Explanation:

Given that:

Pitch (P) = 5 teeth/in

Pressure angle (\Phi) = 20°

Pinion speed (n_p ) = 2000 rev/min

Power (H) = 30 hp

Teeth on gear (N_G) = 50

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Face width (F) = 1 in

Let us first determine the diameter (d) of the pinion.

Diameter (d) = \frac{N}{P}

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= 4 in

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To find the velocity (V); we use the formula:

V = \frac{\pi d n_p}{12}

V = \frac{\pi *4*2000}{12}

V = 2094.40 ft/min

For cut or milled profile; the velocity factor (K_v) can be determined as follows:

K_v = \frac{2000+V}{2000}

K_v = \frac{2000+2094.40}{2000}

= 2.0472

However, there is need to get the value of the tangential load(W^t), in order to achieve that, we have the following expression

W^t=\frac{T}{\frac{d}{2} }

W^t = \frac{63025*H}{\frac{n_pd}{2}}

W^t = \frac{63025*30}{2000*\frac{4}{2}}

W^t = 472.69 lbf

Finally, the bending stress is calculated via the formula:

\sigma = \frac{K_vW^tp}{FY}

\sigma = \frac{2.0472*472.69*5}{1*0.321}

\sigma = 15073.07 psi

\sigma = 15.07 ksi

∴ The estimate of the bending stress = 15.07 ksi

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What is the wavelength of a wave whose speed and period are 72.0 m/s and 5.00 s , respectively ?
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Answer:

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<em>Given that:</em>

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