Understanding Wavelength and Frequency - Physics

Card 0 of 88

Question

A microwave has a wavelength of . What is the frequency of the wave?

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

Answer

The velocity of a wave is equal to the product of the wavelength and frequency:

$v=\lambda f$

We can rearrange this formula to solve for the frequency.

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

Since microwaves are on the electromagnetic spectrum, their velocity will be equal to the speed of light. We are given the wavelength. Using these values, we are able to solve for the frequency.

$f=\frac{310^8\frac{m}{s}}{0.01m}$

$f=310^{10}Hz$

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Question

After exercising, Jane takes her pulse. She realizes that her heart is beating rapidly, approximate four beats every second. What is the period of her elevated heart rate?

Answer

When you see a relationship like "times every second" or "once per hour," these are hints you are looking at a frequency. Frequency is, effectively, how often something happens. If it happens four times per second, then we know how often it happens. The units "per second" are equivalent to Hertz.

The relationship between frequency and period is $T=\frac{1}{f}$ .

Since our given frequency was four beats per second, or , we can solve for the period.

$T=\frac{1}{4Hz}=\frac{1}{4}s=0.25s$

This means that her heart beats once every 0.25 seconds.

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Question

A radio station broadcasts at a frequency of $\small 99,500,000Hz$ . If the broadcast is an electromagnetic wave, then what is its wavelength?

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

Answer

The relationship between wavelength and frequency is given by the equation $\lambda =\frac{c}{f}$ , where $\lambda$ is the wavelength, is the speed of light, and is frequency.

We are given the values for frequency and the speed of light, allowing us to solve for the wavelength.

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

$\lambda=\frac{3*10^8\frac{m}{s}}{99,500,000Hz}$

$\lambda=3.02m$

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Question

A wave oscillates with a speed of $8.9\frac{m}{s}$ and has a wavelength of . What is the frequency of the wave?

Answer

The equation for velocity in terms of wavelength and frequency is $v=\lambda f$ .

We are given the velocity and the wavelength. Using these values, we can solve for the frequency.

$v=\lambda f$

$8.9\frac{m}{s}=6.7m* f$

$\frac{8.9\frac{m}{s}}{6.7m}=f$

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Question

A symphony tunes to an oboe playing a note at . If the speed of sound is $340.29\frac{m}{s}$ , what is the wavelength of this note?

Answer

The relationship between velocity, frequency, and wavelength is:

$v=f\lambda$

Plug in the given information to solve:

$v=f\lambda$

$340.29\frac{m}{s}=440Hz*\lambda$

$\frac{340.29\frac{m}{s}}{440Hz}=\lambda$

$0.773m=\lambda$

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Question

A note is played with a wavelength of . If the speed of sound is $340.29\frac{m}{s}$ , what is the frequency of the note?

Answer

The relationship between velocity, frequency, and wavelength is:

$v=f\lambda$

Plug in the given information to solve:

$v=f\lambda$

$340.29\frac{m}{s}=f*0.444m$

$\frac{340.29\frac{m}{s}}{0.444m}=f$

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Question

A wave with a constant velocity doubles its frequency. What happens to its wavelength?

Answer

The relationship between velocity, frequency, and wavelength is:

$v=f\lambda$

In this case we're given a scenario where $v=f\lambda_1$ and $v=2f*\lambda_2$ . The velocities equal each other because the problem states it has a constant velocity. Therefore we can set these equations equal to each other:

$f\lambda_1=2f\lambda_2$

Notice that the 's cancel out:

$\lambda_1=2\lambda_2$

Divide both sides by two:

$\frac{1}{2}\lambda_1=\lambda_2$

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Question

A wave has a frequency of . What is its period?

Answer

The relationship between frequency and period is $T=\frac{1}{f}$ .

Plug in our given value:

$T=\frac{1}{f}$

$T=\frac{1}{383Hz}$

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Question

A drummer beats his drum every $\frac{1}{6}s$ . What is the frequency of the drum beat?

Answer

Period describes the amount of time per event cycle. The relationship between period and frequency is $f=\frac{1}T$ .

We are given the period, allowing us to solve for the frequency.

$f=\frac{1}T$

$f=\frac{1}{\frac{1}{6}s}$

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Question

A certain type of radiation on the electromagnetic spectrum has a frequency of $10^{20}Hz$ . What is the wavelength of this radiation?

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

Answer

The velocity of a wave is equal to the product of the wavelength and frequency:

$v=\lambda f$

We can rearrange this formula to solve for the wavelength.

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

We can use the given values for the frequency and velocity to solve for the wavelength.

$\lambda=\frac{310^8\frac{m}{s}}{10^{20}Hz}$

$\lambda=310^{-12}m$

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Question

A certain type of radiation on the electromagnetic spectrum has a period of $10^{-17}s$ . What is the wavelength of this radiation?

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

Answer

The velocity of a wave is equal to the product of the wavelength and frequency:

$v=\lambda f$

We can rearrange this formula to solve for the wavelength.

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

We also know that the period is the inverse of the frequency:

$T=\frac{1}{f}$

Substitute this into the equation for wavelength.

$\lambda=cT$

Now we can use our given values for the period and the velocity to solve for the wavelength.

$\lambda=(310^8\frac{m}{s})(10^{-17}s)$

$\lambda=310^{-9}m$

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Question

A certain type of radiation on the electromagnetic spectrum has a period of $10^{-9}s$ . What is the wavelength of this radiation?

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

Answer

The velocity of a wave is equal to the product of the wavelength and frequency:

$v=\lambda f$

We can rearrange this formula to solve for the wavelength.

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

We also know that the period is the inverse of the frequency:

$T=\frac{1}{f}$

Substitute this into the equation for wavelength.

$\lambda=cT$

Now we can use our given values for the period and the velocity to solve for the wavelength.

$\lambda=(310^8\frac{m}{s})*(10^{-9}s)$

$\lambda=0.3m$

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Question

A certain type of radiation on the electromagnetic spectrum has a period of $10^{-9}s$ . What is the frequency of this radiation?

Answer

The relationship between frequency and period is:

$f=\frac{1}{T}$

We are given the period in the question. Using this value, we can solve for the frequency.

$f=\frac{1}{10^{-9}s}$

$f=(10^{-9}s)^{-1}$

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Question

Two notes are played simultaneously. One of them has a period of and the other has a period of . Which one has a longer wavelength?

Answer

The relationship between frequency and wavelength determines the velocity:

$v=\lambda f$

The frequency is the inverse of the period. We can substitute this into the equation above.

$v=\lambda \frac{1}{T}$

In the question, both of the notes are played at the same time in the same location, so they both should have the same velocity. We can set the equation for each tone equal to each other.

$\lambda_1\frac{1}{T_1}=\lambda_2\frac{1}{T_2}$

We are told that . Substitute into our equation.

$\lambda_1\frac{1}{T_1}=\lambda_2\frac{1}{2T_1}$

We can cancel the period from each side of the equation, leaving the relationship between the two wavelengths.

$\lambda_1\frac{1}{T_1}=\lambda_2\frac{1}{2}\frac{1}{T_1}$

$\lambda_1=\frac{1}{2}\lambda_2$

The wavelength of the first wave is equal to half the wavelength of the second. This means that the wavelength for the tone with a longer period will have a longer wavelength as well.

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Question

A clarinet plays a note with a frequency of . What is the wavelength of the sound?

$v_s=340.29\frac{m}{s}$

Answer

The relationship between wavelength, frequency, and velocity is:

$v=f\lambda$

We are given the speed of sound and the frequency, allowing us to solve for the wavelength.

$340.29\frac{m}{s}=(466.16Hz)\lambda$

$\lambda=\frac{340.29\frac{m}{s}}{466.16Hz}$

$\lambda=0.73m$

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Question

A saxophone plays a note with a wavelength of . What is the frequency?

$v_s=340.29\frac{m}{s}$

Answer

The relationship between wavelength, frequency, and velocity is:

$v=f\lambda$

We are given the speed of sound and the wavelength, allowing us to solve for the frequency.

$340.29\frac{m}{s}=f(0.88m)$

$f=\frac{340.29\frac{m}{s}}{0.88m}$

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Question

What is the period of a wave with a frequency of ?

Answer

The relationship between frequency and period is:

$T=\frac{1}{f}$

Period is simply the reciprocal of frequency.

$T=\frac{1}{61Hz}$

$T=\frac{1}{61}s$

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Question

A cat meows with a frequency of . What is the wavelength of this sound?

$v_{sound}=340.29\frac{m}{s}$

Answer

The relationship between velocity, wavelength, and frequency is:

$v=f\lambda$

We know the speed of sound and the frequency of the sound, allowing us to solve for the wavelength.

$340.29\frac{m}{s}=(512Hz)\lambda$

$\lambda=\frac{340.29\frac{m}{s}}{512Hz}$

$\lambda=0.66m$

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Question

A wave has a period of . What is the frequency of this wave?

Answer

Frequency is equal to the reciprocal of the period:

$f=\frac{1}{T}$

Given the period, we can invert it to find the frequency.

$f=\frac{1}{3s}$

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Question

The period of a wave is . What is the frequency?

Answer

Frequency is the reciprocal of period.

$f=\frac{1}{T}$

We are given the period, so we simply need to take the reciprocal to solve for the frequency.

$f=\frac{1}{13s}$

$f=\frac{1}{13}Hz$

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