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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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My favorite playground equipment is the swings.

4 0
3 years ago
A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t
Yuliya22 [10]

Answer:

a) W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

b) W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

c) V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

d)  d_1 =0.183m or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

Part b

For this case first we can convert the spring constant to N/m like this:

2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}

And the work donde by the spring on this case is given by:

W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i

And if we solve for the initial velocity we got:

V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

-1/2mV^2_i = W_g + W_{spring}

And replacing we got:

-1/2mV^2_i =mg d_1 -1/2 k d^2_1

And we can put the terms like this:

\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0

If we multiply all the equation by 2 we got:

k d^2_1 -2 mg d_1 -m V^2_i =0

Now we can replace the values and we got:

200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0

200 d^2_1 -2.058 d_1 -6.3525=0

And solving the quadratic equation we got that the solution for d_1 =0.183m or 18.3 cm because the negative solution not make sense.

5 0
3 years ago
The frequency of a wave is 200 Hz. The wavelength is 0.1 m. What is the period of the wave?
Aleks04 [339]
The formula for the period of wave is: wave period is equals to 1 over the frequency.waveperiod=\frac{1}{frequency}
To get the value of period of wave you need to divide 1 by 200 Hz. However, beforehand, you have to convert 200 Hz to cycles per second. So that would be, 200 cyles per second or 200/s.
By then, you can start the computation by dividing 1 by 200/s. Since 200/s is in fractional form, you have to find its reciprocal form and multiply it to one which would give you 1 (one) second over 200. This would then lead us to the value 0.005 seconds as the wave period.

wave period= 1/200 Hz
Convert Hz to cycles per second first
200 Hz x 1/s= 200/second
Make 200/second as your divisor, so:

wave period= 1/ 200/s

get the reciprocal form of 200/s which is s/200

then you can start the actual computation:

wave period= 1 x s divided by 200

this would give us an answer of 0.005 s. 
6 0
3 years ago
The glass in a window is 35 inches wide and 20 inches tall, and standard atmospheric pressure is 14.7 pounds per square inch. Wh
iogann1982 [59]

Answer:103 pounds

Explanation:

Given

width of window b=35 in.

height of window h=20 in.

standard atmospheric pressure P_{outside}=14.7 psi

Also P_{inside}=1.01P_{outside}

Thus Net Force on the window will be the algebraic sum of Force due to outside and inside Pressure .

F_{net}=(P_{inside}-P_{outside})\cdot A

F_{net}=P_{outside}(1.01-1)\times 35\times 20

F_{net}=14.7\times 0.01\times 35\times 20

F_{net}=102.9\ pounds\approx 103\ pounds            

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B. Movement would be the correct answer.
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3 years ago
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