# Bouncing ball and Doppler effect

• Soren4
In summary: No, the ball emits continuously at frequency fo=430 Hz. The emitted sound makes sound waves traveling in air at higher frequency in the direction of the velocity of the ball, and at lower frequency in the opposite direction. The sound emitted by the ball is connected to the vibration of the ball or some part of it, but the sound wave traveling in the air is connected to vibration of air molecules. It is the sound wave traveling in air that reflects from the walls and interfere with the sound wave traveling in opposite direction. Maybe that's a wrong complication to the problem that I'm thinking about, but is it really only necessary to use doppler effect and neglect the fact that the ball is continuously emitting? Yes, that's right
Soren4

## Homework Statement

Consider two parallel walls perfectly reflective placed at the distance ##d = 0.8 m ##. A ball, provided with a device through which are emitted continuously frequency sound waves equal to ##f_0=430 Hz##, is launched from one wall to another. It moves with constant velocity ##v##. After some time, an observer placed at the position where the ball was thrown a beat frequency equal to ##5 Hz##.
Compute:

a) the speed of the ball;

b) assuming the formation of standing waves when the two walls are made the same traveling conditions (nodes) of the gas particles, determine the speed at which must travel the ball so that between the two walls standing waves will be generated with frequency equal to twice the fundamental frequency. Neglect, in this calculation, the higher frequency sound wave presence.

Assume the speed of sound ##c = 340 m / s##.

[Results: a) ## v = 1.98 m / s ##; b) ## v = 4 m / s ##]

## Homework Equations

Doppler effect : $$f^*=f_0(\frac{c}{c\pm v_{ball}})$$

## The Attempt at a Solution

a. I get confused because the result is correct if I use
$$f_{beats}=|f_0(\frac{c}{c- v_{ball}})-f_0(\frac{c}{c+v_{ball}})|$$
But this does not seem correct to me. Immediately after the first bounce the ball i moving towards the observer, so it should be
$$f_{beats}=|f_0(\frac{c}{c- v_{ball}})-f_0(\frac{c}{c-v_{ball}})|=0$$

Which of course is not.

b. This is the point I cannot do at all. What is the strategy to use here?

Any help is really appreciated.

Soren4 said:

## Homework Statement

Consider two parallel walls perfectly reflective placed at the distance ##d = 0.8 m ##. A ball, provided with a device through which are emitted continuously frequency sound waves equal to ##f_0=430 Hz##, is launched from one wall to another. It moves with constant velocity ##v##. After some time, an observer placed at the position where the ball was thrown a beat frequency equal to ##5 Hz##.
Compute:

a) the speed of the ball;

b) assuming the formation of standing waves when the two walls are made the same traveling conditions (nodes) of the gas particles, determine the speed at which must travel the ball so that between the two walls standing waves will be generated with frequency equal to twice the fundamental frequency. Neglect, in this calculation, the higher frequency sound wave presence.

Assume the speed of sound ##c = 340 m / s##.

[Results: a) ## v = 1.98 m / s ##; b) ## v = 4 m / s ##]

## Homework Equations

Doppler effect : $$f^*=f_0(\frac{c}{c\pm v_{ball}})$$

## The Attempt at a Solution

a. I get confused because the result is correct if I use
$$f_{beats}=|f_0(\frac{c}{c- v_{ball}})-f_0(\frac{c}{c+v_{ball}})|$$
But this does not seem correct to me. Immediately after the first bounce the ball i moving towards the observer, so it should be
$$f_{beats}=|f_0(\frac{c}{c- v_{ball}})-f_0(\frac{c}{c-v_{ball}})|=0$$

Which of course is not.

The ball continuously emits sound. The sound also reflects from the wall, so the observer hears the sounds emitted by both the receding and approaching balls.

Soren4
ehild said:
The ball continuously emits sound. The sound also reflects from the wall, so the observer hears the sounds emitted by both the receding and approaching balls.

I thought about the problem for a while, I'm totally ok with point a. now, but point b. is still a mistery for me.

How does the speed of the ball influences the formation of standing waves?

The frequency twice of the fundamental is ##f=2 \frac{c}{2d}## but the ball emits at ##f_0## regardless its velocity.

So can I deal with point b?

The
Soren4 said:
How does the speed of the ball influences the formation of standing waves?

The frequency twice of the fundamental is ##f=2 \frac{c}{2d}## but the ball emits at ##f_0## regardless its velocity.

So can I deal with point b?
The speed of the ball influences the frequency of the sound waves which propagate in air forward and backward between the walls. These two waves generate the standing wave. The problem says to neglect the higher frequency wave. So you have sound wave of frequency f*=f0 c/(c+v), find v.

Soren4
ehild said:
The

The speed of the ball influences the frequency of the sound waves which propagate in air forward and backward between the walls. These two waves generate the standing wave. The problem says to neglect the higher frequency wave. So you have sound wave of frequency f*=f0 c/(c+v), find v.

Thanks for the reply! I thought about it again and I understood your solution, except one thing.

The ball is continuosly emitting, so the reflected waves interfere not only with the other reflected waves, but also with the waves just emitted by the moving ball, don't they? So shouldn't this be taken into account? I mean the ball should have a velocity such that it does not emit at all when it is in the position of a node and emits when it is on an antinode?

Maybe that's a wrong complication to the problem that I'm thinking about, but is it really only necessary to use doppler effect and neglect the fact that the ball is continously emitting?

Soren4 said:
Thanks for the reply! I thought about it again and I understood your solution, except one thing.

The ball is continuosly emitting, so the reflected waves interfere not only with the other reflected waves, but also with the waves just emitted by the moving ball, don't they? So shouldn't this be taken into account? I mean the ball should have a velocity such that it does not emit at all when it is in the position of a node and emits when it is on an antinode?
No, the ball emits continuously at frequency fo=430 Hz. The emitted sound makes sound waves traveling in air at higher frequency in the direction of the velocity of the ball, and at lower frequency in the opposite direction. The sound emitted by the ball is connected to the vibration of the ball or some part of it, but the sound wave traveling in the air is connected to vibration of air molecules.
It is the sound wave traveling in air that reflects from the walls and interfere with the sound wave traveling in opposite direction.
Soren4 said:
Maybe that's a wrong complication to the problem that I'm thinking about, but is it really only necessary to use doppler effect and neglect the fact that the ball is continously emitting?
Yes, it would be enough to consider only the sound wave emitted at an instant, and ignoring the emitting ball. But that wave would extinct in a short time. The continuous emission of sound ensures constant intensity, making the formation of standing waves possible.
There are sound waves with two different frequencies, but the problem says to ignore the one with higher frequency. And the conditions correspond to standing wave formation for one frequency only.

Soren4

## 1. How does the Doppler effect affect the sound of a bouncing ball?

The Doppler effect is the change in frequency of a sound wave due to the relative motion of the source and the observer. When a ball bounces, it creates sound waves that travel to our ears. As the ball moves towards us, the sound waves are compressed, resulting in a higher frequency and a higher pitch. As the ball moves away from us, the sound waves are stretched, resulting in a lower frequency and a lower pitch.

## 2. Why does a bouncing ball make a different sound when it bounces on different surfaces?

The material and texture of the surface that the ball bounces on can affect the sound it produces. For example, a hard surface like concrete will produce a sharper and louder sound compared to a softer surface like grass. This is because different surfaces absorb and reflect sound waves differently, resulting in a variation of the sound produced by the bouncing ball.

## 3. How does the height of a bounce affect the Doppler effect?

The height of the bounce does not directly affect the Doppler effect. However, a higher bounce can result in a longer distance traveled by the ball, which can then affect the intensity and duration of the sound waves produced. This can in turn affect the perceived pitch of the sound due to the Doppler effect.

## 4. Can the Doppler effect be observed with a stationary bouncing ball?

No, the Doppler effect can only be observed when there is relative motion between the source of the sound (the bouncing ball) and the observer (our ears). If the ball is stationary, there is no change in frequency and therefore no Doppler effect.

## 5. How is the Doppler effect used in real-life applications?

The Doppler effect has many practical applications, such as in weather forecasting, radar systems, and medical imaging. It is also used in police radar guns to measure the speed of moving vehicles. In astronomy, the Doppler effect is used to measure the speed and distance of celestial objects based on the shift in their light waves.

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