How Do Temperature Changes Affect Ultrasonic Distance Measurements?

[JSapit]

Homework Statement

The ultrasonic motion sensor sends pulses of the ultrasound toward a cart on the low-friction track and determines the distance by the time an echo takes to return. The temperature in the lab is equal to 20 °C. What is the distance between the motion sensor and the cart, if the reflected echo was recorded after 5.33 ms (1 ms = 0.001 s)?

(a) The distance is = (meters)

The velocity of sound in air is a function of temperature and changes by 0.6 m/s for every degree Celsius (°C) of the temperature change. The velocity of sound is proportional to the temperature. What would be the apparent distance between the motion sensor and the cart, if the temperature in the lab increases by DT = 12.74 °C?


(b) Apparent distance for the increased temperature is = (in meters)

3.) The following graph is an illustration of the phenomenon called beats. Beats are due to a superposition of two wavs with similar amplitudes, but slightly different frequencies. From the graph below, find the approximate value of the beat frequency. Note that one division on the vertical axis corresponds to 0.050 V, whereas one division on the time (horizontal axis) corresponds to 10 ms.

https://chip.physics.purdue.edu/protected/Prelab220newimg/m10q3beats.jpg"

(a) The beat frequency is = (in Hz)

They give us that the speed of sound in air at 20 degrees C is 343 m/s.

Homework Equations

V=D/t, f=1/T, T=1/f

The Attempt at a Solution

1.) I tried converting 5.33 ms to seconds, and the multiplying it by the velocity to get the distance, but that was wrong. I also tried multiplying the distance I found by two, for no apparent reason, but that was wrong two. The lab book doesn't give us any more equations as to how to figure it out, so I'm confused.

3.) I'm not sure how to find the beat frequency.

 

[JSapit]

I've been working on this all day and I can't get it. Does anyone have any idea?

 

[Snazzy]

For part A, you say you multiplied the number that you obtained by 2. Why? The reflected echo was recorded after 5.33 seconds. So that must mean the reflected echo got to the cart, and then came back.

For part B, the beat frequency is given by $$f_{beat}=|f_2-f_1|$$

 

[JSapit]

For part A, I didn't know what else to do for it, so I thought why not. How would I go about solving this? Is there a specific formula for this scenario?

For number 3, I know about the beat formula, but I'm having trouble interpreting the graph. How did you interpret it?

 

[Snazzy]

The time for the echo to go to the cart and come back again is 5.33 ms. Draw a picture.

You have two superimposed beats on top of each other. You've got a big one with a lot of little ones inside the big ones. Find the frequency of each.

 

[JSapit]

Okay, I got the one with the beats. But I'm still confused with the question with the echo. What equation do I use to solve for it? I used V=d/t and solved for d, which is how I got those answers I mentioned above, but they're wrong. Is there another equation I have to use?

Thanks for your help.

 

[Snazzy]

Okay yes, using v = d/t is the right equation.

But when solving for d, think what the d represents. It takes 5.33 ms to go from the sensor, to the cart, back to the sensor.

 

[JSapit]

So would I multiply that by three? I'm still confused, sorry.

 

[JSapit]

Nevermind. I got it. Thank you.