Understanding Wavelength and Frequency

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Physics › Understanding Wavelength and Frequency

Questions 1 - 10

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

Explanation

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.

$v=\lambda f$

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

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

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

The new wavelength will be $\frac{1}{2}$ the old wavelength.

The new wavelength will also double.

The wavelengths will be the same.

There is insufficient information to solve.

Explanation

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:

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

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?

Explanation

The relationship between velocity, frequency, and wavelength is:

$v=f\lambda$

Plug in the given information to solve:

$v=f\lambda$

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

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

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?

Explanation

The relationship between velocity, frequency, and wavelength is:

$v=f\lambda$

Plug in the given information to solve:

$v=f\lambda$

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

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

$0.773m=\lambda$

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

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

$3*10^{17}m$

$3*10^{-17}m$

$3*10^{1.125}m$

Explanation

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.

$\lambda=c*T$

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

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

$\lambda=0.3m$

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

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

$3*10^{10}Hz$

$3*10^{-11}Hz$

$3*10^{-6}Hz$

Explanation

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.

$f=\frac{3*10^8\frac{m}{s}}{0.01m}$