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morpeh [17]
2 years ago
14

A series of pulses of amplitude 0.28 m are sent down a string that is attached to a post at one end. The pulses are reflected at

the post and travel back along the string without loss of amplitude. What is the amplitude at a point on the string where two pulses cross if the string is rigidly attached to the post?
Physics
2 answers:
andreyandreev [35.5K]2 years ago
5 0

Answer:

The answer is zero displacement (0 m)

Explanation:

If the end of the string is not free, the reflected pulse has the same amplitude but opposite polarity to the incident pulse. Because of this, for A, the result equals zero (0 m).

irina1246 [14]2 years ago
3 0

Answer:

a. The pulses cancel each other resulting in zero displacement.

b. The pulses reinforce each other, having a displacement of 2× amplitude or 0.56 m.

Explanation:

When the pulse are sent down the attached string  it gets reflected and we have crossing pulses as the incident and reflected pulses cross each other.

a. If the string is rigidly attached to the post then the incident and reflected pulses will have the same amplitude but different direction. That is either the reflected will be going up and the incident down thereby resulting in a cancellation or zero displacement

Incident amplitude = 0.28 m

Reflected amplitude = -0.28 m

Sum = 0.28 m - 0.28 m = 0 m

b. If the end at which the relfection occurs is free to slide, then the incident and the reflected pulses will again have the same amplitude and in this case, the same direction. Therefore;

Incident amplitude = 0.28 m

Reflected amplitude = 0.28 m

Sum = 0.28 m + 0.28 m = 0.56 m

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Answer:

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In terms of displacement, average speed is 0 km/h.

Explanation:

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= (40.0 - 10.0) + (20.0 - 10.0) + (40.0 - 20.0) \\  = 30.0 + 10.0 + 20.0 \\  = 60.0 \: km

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average \: speed =  \frac{total \: distance}{total \: time}  \\

In terms of distance.

substitute:

average \: speed =  \frac{60.0}{3.0}  \\  \\  = 20 \:  {kmh}^{ - 1}

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as the particle moves perpendicular to the field, the angle is θ= 90º

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The exercise indicates the time it takes in the route that is carried out with constant speed, therefore we can use

          v = d / t

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          d = \frac{1}{4} \  2\pi  r

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we substitute

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Let's reduce to mC

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