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AP Physics 1 : Waves and Velocity

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Example Questions

Example Question #1 : Waves And Velocity

You and your friends are spending an afternoon at a lake. You are racing a friend across the lake while staying under the water. Being the faster swimmer, you are averaging a pace of about $1.5\frac{m}{s}$ . Another friend on shore is cheering you on. If the average frequency of your friend's voice is 500Hz, and the wavelength of the sound in the water is 0.85m, what is the speed of the sound in the water?

$v_{s_{air}}=340\frac{m}{s}$

Possible Answers:

$340\frac{m}{s}$

$556\frac{m}{s}$

$425\frac{m}{s}$

$400\frac{m}{s}$

$588\frac{m}{s}$

Correct answer:

$425\frac{m}{s}$

Explanation:

This problem emphasizes the importance of being able to sift through a problem statement and pick out only the information you need to solve the problem. We are given four values, but only need two of them. If you are unable to sift through the problem effectively, you are likely to get bogged down by all the details and waste time performing unneeded calculations.

To find the speed of sound in the water, we only need one equation:

$v = f\lambda$

We are given the frequency of the sound and its wavelength in water. It is important to note that the frequency of a sound does not change as it enters a different medium; only the velocity and wavelength do.

Plugging in our values, we get:

$v = 500 Hz \cdot 0.85 m = 425 \frac{m}{s}$

If you got the answer 400 m/s, watch your units! You probably divided 340 m/s by 0.85 m, which gives you units of 1/s. That is a unit of frequency, not velocity.

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Example Question #1 : Waves And Velocity

One of your friends decides to do a massive cannonball into a swimming pool in an attempt to splash the lifeguard. Your friend hits the water from where the lifeguard is and the cannonball creates a series of waves. If one wave is created every 2 seconds, and there are four full waves between your friend and the lifeguard, what is the speed of the waves created?

Possible Answers:

$0.63\frac{m}{s}$

$0.40 \frac{m}{s}$

$1.67 \frac{m}{s}$

$0.12\frac{m}{s}$

$2.5\frac{m}{s}$

Correct answer:

$0.63\frac{m}{s}$

Explanation:

In order to determine the velocity of a wave, we need to know the frequency and wavelength. From the problem statement, we know that a wave is created every 2 seconds. This is the period of the wave; therefore we can say that the frequency is:

$f = \frac{1}{T}=\frac{1}{2s}=0.5Hz$

Now we just need the wavelength. We are told that there are four full waves over a distance of 5 meters.

$\lambda = \frac{5m}{4\ \text{waves}} = 1.25m$

We can use the following formula to get the velocity:

$v = f\lambda = (0.5Hz)(1.25m) = 0.63\frac{m}{s}$

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Example Question #3 : Waves And Velocity

A major earthquake about 600km off the coast of California occurs at 9:28 AM. If you are 800km from the epicenter and feel the earthquake at 10:43 AM, how fast are the waves traveling?

Possible Answers:

$275\frac{m}{s}$

$178\frac{m}{s}$

$586\frac{m}{s}$

$414\frac{m}{s}$

$334\frac{m}{s}$

Correct answer:

$178\frac{m}{s}$

Explanation:

To determine the velocity of the created waves, we need to know how quickly they cover a certain distance. The problem statement tells us that the covered distance is 800km. From the given time differences, we know that it takes 1 hour and 15 minutes to cover that distance. We will need to convert this to seconds:

$t = 75 min \left ( \frac{60s}{min}\right ) = 4500 s$

Now we can calculate the speed of the waves:

$v = \frac{d}{t}=\frac{800,000m}{4500 s} = 178\frac{m}{s}$

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Example Question #1 : Waves And Velocity

On a piano, the note known as "middle C" has a frequency of roughly 262 hertz.

The coldest temperature ever recorded on Earth was $-89^{\circ}C$ . At this temperature, sound travels at $272\frac{m}{s}$ in air.

If Carol plays middle C on a piano at this temperature, what will the sound's wavelength be?

Possible Answers:

1.28m

1.72m

1.40m

1.04 m

0.96 m

Correct answer:

1.04 m

Explanation:

For any wave,

$v=\lambda f$

We know that the frequency of middle C is 262 hertz, and we also know that the speed of sound is $272\frac{m}{s}$ . Therefore:

$272 \frac{m}{s}=\lambda(262s^{-1})$

Solving for wavelength yields

$\lambda = 1.04m$

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Example Question #5 : Waves And Velocity

A light wave has a wavelength of 0.5m . What is its frequency?

$c=3\cdot10^8\frac{m}{s}$

Possible Answers:

$1.5\cdot 10^4Hz$

$2\cdot 10^9Hz$

$6\cdot 10^8Hz$

$1.5\cdot 10^8Hz$

10^7Hz

Correct answer:

$6\cdot 10^8Hz$

Explanation:

The equation relaing velocity, wavelength, and frequency is:

$v = \lambda f$ .

Solving for frequency,

$3\cdot10^8\frac{m}{s} = 0.5m\cdot f$

$f = 6\cdot10^8 Hz$

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Example Question #1201 : Ap Physics 1

If we wanted to double wave velocity, but could only change frequency and wavelength by the same factor, by what factor would we have to change both frequency and wavelength?

Possible Answers:

We would have to change both frequency and wavelength by a factor of $\sqrt 2$

We would have to change both frequency and wavelength by a factor of $\frac{1}{2}$

We would have to change both frequency and wavelength by a factor of

Correct answer:

We would have to change both frequency and wavelength by a factor of $\sqrt 2$

Explanation:

Wave speed is related to wavelength and frequency by:

$v=\lambda f$ , where $\lambda$ is wavelength and is frequency.

If we wanted to change by 2, and change both $\lambda$ and by the same factor , we could set this up as:

$2v=a\lambda *af$

If we isolated just the factors by which we're changing the wave speed,

2=a^2

$\sqrt 2=a$

We'd have to change both wavelength and frequency by $\sqrt 2$

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Example Question #7 : Waves And Velocity

What is the speed of light when passed through glass with an index of refraction of 2?

$c=3*10^8\frac{m}{s}$

Possible Answers:

$1.5*10^8 \frac{m}{s}$

$1.0*10^8 \frac{m}{s}$

$3.0*10^8 \frac{m}{s}$

$6.0*10^7 \frac{m}{s}$

Correct answer:

$1.5*10^8 \frac{m}{s}$

Explanation:

The equation for the speed of light through materials of different indices of refraction is given by the following equation:

$n=\frac{c}{v}$

Solve for velocity:

$v= \frac{3*10^8\frac{m}{s}}{2}= 1.5*10^8\frac{m}{s}$

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Example Question #8 : Waves And Velocity

waves are hitting the side of the pier in a regular fashion. waves hit the pier during 60s , and the waves are between crests, calculate the velocity of the waves.

Possible Answers:

$0.58 \frac{m}{s}$

$1.2\frac{m}{s}$

$4 \frac{m}{s}$

$2.3 \frac{m}{s}$

Correct answer:

$2.3 \frac{m}{s}$

Explanation:

We can calculate the velocity of a wave if we know the wavelength $\lambda$ and the frequency of the wave by the following equation,

$v = \lambda f$

We are told $\lambda = 4m$ , and can calculate the frequency

$f = \frac{35}{60s}=0.583 \frac{1}{s}$

Therefore, we can find the velocity of the waves,

$v = \lambda f = (4m)0.583\frac{1}{s}=2.3 \frac{m}{s}$

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Example Question #1 : Waves And Velocity

A wave emitter is set to a frequency of $2450\textup{ MHz}$ . It is found that there is $6.0\textup{ cm}$ between anti-nodes of the waves created by this instrument. What is the speed of the waves as measured by this experiment?

Possible Answers:

$4.08*10^{10}\ \frac{\textup{m}}{\textup{s}}$

$1.47*10^{8}\ \frac{\textup{m}}{\textup{s}}$

$8.17*10^{10}\ \frac{\textup{m}}{\textup{s}}$

$3.00*10^{8}\ \frac{\textup{m}}{\textup{s}}$

$2.94*10^{8}\ \frac{\textup{m}}{\textup{s}}$

Correct answer:

$2.94*10^{8}\ \frac{\textup{m}}{\textup{s}}$

Explanation:

The distance between each anti-node in a standing wave is only half a wavelength, so we need to multiply the distance measured between anti-nodes by 2. Multiply the wavelength by the frequency to get the speed of the wave.

$v=\lambda f$

We need to remember to convert our centimeters to meters and our megaHertz to Hertz so the units work out:

$v=(2.450*10^{9}Hz)(.12m)$

$v= 2.94x10^{8}\frac{m}{s}$

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Example Question #51 : Electricity And Waves

Determine the wavelength of a wave traveling through a material with tension 10N, a linear mass density of $\mu= 35\frac{kg}{m}$ , and frequency 10Hz.

Possible Answers:

$\frac{1}{70}m$

None of these

0.053m

0.53m

Correct answer:

0.053m

Explanation:

We need to utilize two equations for this question. The first of which is:

$v=\sqrt{\frac{T}{\mu}}$

Here, is wave speed, is tension force of the material, and $\mu$ is linear density. The othe equation we need to use is:

$v=\lambda f$

Here, is wave speed, $\lambda$ is wavelength and is frequency.

In our case:

$T=10N, \mu=35 \frac{kg}{m}$

plug in known values and solve for wave speed.

$v=\sqrt{\frac{10N}{35\frac{kg}{m}}}=0.5345\frac{m}{s}$

Solve the second equation for $\lambda$ .

$\lambda=\frac{v}{f}$

Plug in known values and solve for wavelength.

$\lambda=\frac{0.5345\frac{m}{s}}{10s^{-1}}=0.053m$

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AP Physics 1 : Waves and Velocity

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #1 : Waves And Velocity

Example Question #1 : Waves And Velocity

Example Question #3 : Waves And Velocity

Example Question #1 : Waves And Velocity

Example Question #5 : Waves And Velocity

Example Question #1201 : Ap Physics 1

Example Question #7 : Waves And Velocity

Example Question #8 : Waves And Velocity

Example Question #1 : Waves And Velocity

Example Question #51 : Electricity And Waves

All AP Physics 1 Resources

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