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High School Physics : Understanding Wavelength and Frequency

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Example Questions

Example Question #21 : Waves

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

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

Possible Answers:

$3*10^{-25}m$ $\displaystyle 3*10^{-25}m$

$3*10^{-17}m$ $\displaystyle 3*10^{-17}m$

$3*10^{-2.125}m$ $\displaystyle 3*10^{-2.125}m$

$3*10^{8}m$ $\displaystyle 3*10^{8}m$

$3*10^{-9}m$ $\displaystyle 3*10^{-9}m$

Correct answer:

$3*10^{-9}m$ $\displaystyle 3*10^{-9}m$

Explanation:

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

$v=\lambda f$ $\displaystyle v=\lambda f$

We can rearrange this formula to solve for the wavelength.

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

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

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

Substitute this into the equation for wavelength.

$\lambda=c*T$ $\displaystyle \lambda=c*T$

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

$\lambda=(3*10^8\frac{m}{s})*(10^{-17}s)$ $\displaystyle \lambda=(3*10^8\frac{m}{s})*(10^{-17}s)$

$\lambda=3*10^{-9}m$ $\displaystyle \lambda=3*10^{-9}m$

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Example Question #21 : Waves, Sound, And Light

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

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

Possible Answers:

$3*10^{-17}m$ $\displaystyle 3*10^{-17}m$

0.3m $\displaystyle 0.3m$

$3*10^{17}m$ $\displaystyle 3*10^{17}m$

$\displaystyle 3m$

$3*10^{1.125}m$ $\displaystyle 3*10^{1.125}m$

Correct answer:

0.3m $\displaystyle 0.3m$

Explanation:

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

$v=\lambda f$ $\displaystyle v=\lambda f$

We can rearrange this formula to solve for the wavelength.

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

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

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

Substitute this into the equation for wavelength.

$\lambda=c*T$ $\displaystyle \lambda=c*T$

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

$\lambda=(3*10^8\frac{m}{s})*(10^{-9}s)$ $\displaystyle \lambda=(3*10^8\frac{m}{s})*(10^{-9}s)$

$\lambda=0.3m$ $\displaystyle \lambda=0.3m$

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Example Question #22 : Waves, Sound, And Light

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

Possible Answers:

10^9Hz $\displaystyle 10^9Hz$

$10^{17}Hz$ $\displaystyle 10^{17}Hz$

$10^{-9}Hz$ $\displaystyle 10^{-9}Hz$

$10^{18}Hz$ $\displaystyle 10^{18}Hz$

10^3Hz $\displaystyle 10^3Hz$

Correct answer:

10^9Hz $\displaystyle 10^9Hz$

Explanation:

The relationship between frequency and period is:

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

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

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

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

$\displaystyle f=10^9Hz$

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Example Question #23 : Waves, Sound, And Light

Two notes are played simultaneously. One of them has a period of $\displaystyle T$ and the other has a period of $\displaystyle 2T$. Which one has a longer wavelength?

Possible Answers:

They have the same wavelength

$\displaystyle 2T$

We need to know the period in order to solve

We need to know the frequency in order to solve

$\displaystyle T$

Correct answer:

$\displaystyle 2T$

Explanation:

The relationship between frequency and wavelength determines the velocity:

$v=\lambda f$ $\displaystyle v=\lambda f$

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

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

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.

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

We are told that $\displaystyle T_2=2T_1$. Substitute into our equation.

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

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

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

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

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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Example Question #11 : Understanding Wavelength And Frequency

What is the period of a wave with a frequency of 61Hz $\displaystyle 61Hz$?

Possible Answers:

$\frac{1}{61}s$ $\displaystyle \frac{1}{61}s$

61s $\displaystyle 61s$

$\frac{1}{3721}s$ $\displaystyle \frac{1}{3721}s$

3721s $\displaystyle 3721s$

We need to know the wavelength in order to solve

Correct answer:

$\frac{1}{61}s$ $\displaystyle \frac{1}{61}s$

Explanation:

The relationship between frequency and period is:

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

Period is simply the reciprocal of frequency.

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

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

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Example Question #25 : Waves, Sound, And Light

A cat meows with a frequency of 512Hz $\displaystyle 512Hz$. What is the wavelength of this sound?

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

Possible Answers:

171.71m $\displaystyle 171.71m$

There is insufficient information to solve

852.29m $\displaystyle 852.29m$

0.66m $\displaystyle 0.66m$

1.5m $\displaystyle 1.5m$

Correct answer:

0.66m $\displaystyle 0.66m$

Explanation:

The relationship between velocity, wavelength, and frequency is:

$v=f\lambda$ $\displaystyle v=f\lambda$

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

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

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

$\lambda=0.66m$ $\displaystyle \lambda=0.66m$

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Example Question #26 : Waves, Sound, And Light

A saxophone plays a note with a wavelength of 0.88m $\displaystyle 0.88m$. What is the frequency?

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

Possible Answers:

2.91Hz $\displaystyle 2.91Hz$

0.003Hz $\displaystyle 0.003Hz$

216Hz $\displaystyle 216Hz$

$\displaystyle 386.69Hz$

$\displaystyle 343.72Hz$

Correct answer:

$\displaystyle 386.69Hz$

Explanation:

The relationship between wavelength, frequency, and velocity is:

$v=f\lambda$ $\displaystyle v=f\lambda$

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

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

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

$\displaystyle f=386.69Hz$

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Example Question #31 : Waves

A clarinet plays a note with a frequency of $\displaystyle 466.16Hz$. What is the wavelength of the sound?

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

Possible Answers:

125.87m $\displaystyle 125.87m$

158.6m $\displaystyle 158.6m$

0.08m $\displaystyle 0.08m$

0.73m $\displaystyle 0.73m$

1.37m $\displaystyle 1.37m$

Correct answer:

0.73m $\displaystyle 0.73m$

Explanation:

The relationship between wavelength, frequency, and velocity is:

$v=f\lambda$ $\displaystyle v=f\lambda$

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

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

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

$\lambda=0.73m$ $\displaystyle \lambda=0.73m$

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Example Question #31 : Waves

A wave has a period of $\displaystyle 3s$. What is the frequency of this wave?

Possible Answers:

0.33Hz $\displaystyle 0.33Hz$

0.11Hz $\displaystyle 0.11Hz$

9Hz $\displaystyle 9Hz$

3Hz $\displaystyle 3Hz$

Correct answer:

0.33Hz $\displaystyle 0.33Hz$

Explanation:

Frequency is equal to the reciprocal of the period:

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

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

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

$\displaystyle f=0.33Hz$

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Example Question #11 : Understanding Wavelength And Frequency

The period of a wave is 13s $\displaystyle 13s$. What is the frequency?

Possible Answers:

We need to know the velocity of the wave in order to solve

13Hz $\displaystyle 13Hz$

169Hz $\displaystyle 169Hz$

$\frac{1}{13}Hz$ $\displaystyle \frac{1}{13}Hz$

We need to know the wavelength in order to solve

Correct answer:

$\frac{1}{13}Hz$ $\displaystyle \frac{1}{13}Hz$

Explanation:

Frequency is the reciprocal of period.

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

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

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

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

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High School Physics : Understanding Wavelength and Frequency

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #21 : Waves

Example Question #21 : Waves, Sound, And Light

Example Question #22 : Waves, Sound, And Light

Example Question #23 : Waves, Sound, And Light

Example Question #11 : Understanding Wavelength And Frequency

Example Question #25 : Waves, Sound, And Light

Example Question #26 : Waves, Sound, And Light

Example Question #31 : Waves

Example Question #31 : Waves

Example Question #11 : Understanding Wavelength And Frequency

All High School Physics Resources

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