Physics Formulas

Amplitude Formula

Amplitude is something that relates to the maximum displacement of the waves. Furthermore, in this topic, you will learn about the amplitude, amplitude formula, formula’s derivation, and solved example. Besides, after completing the topic you will be able to understand amplitude.

amplitude formula

Amplitude

It refers to maximum displacement from the equilibrium that an object in periodic motion show. As an example, a pendulum swings through its equilibrium point (straight down), and then swing to a maximum distance away from the center.

Furthermore, the distance of the amplitude is A. Moreover, the full range of the pendulum has a magnitude of 2A. Besides, the periodic motion also applies to the waves and springs. In addition, the sine function oscillates between values of +1 and -1, so it is used to describe periodic motion.

Most noteworthy, the unit of amplitude is a meter (m).

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Amplitude Formula

Position = amplitude × sine function (angular frequency × time + phase difference)

x = A sin (\(\omega t + \phi\))

Derivation of the Amplitude Formula

x = refers to the displacement in Meters (m)
A = refers to the amplitude in meters (m)
\(\omega\) = refers to the angular frequency in radians per seconds (radians/s)
t = refers to the time in seconds (s)
\(\phi\) = refers to the phase shift in radians

Solved Examples

Example 1

Assume that a pendulum is swinging back and forth. Also, the angular frequency of the oscillation is \(\omega\) = \(\pi\) radians/s, and the phase shift is \(\phi\) = 0 radians. Moreover, the time t = 8.50 s, and the pendulum is 14.0 cm or x = 0.140 m. So, calculate the amplitude of the oscillation?

Solution:

x = 0.140 m
\(\omega\) = \(\pi\) radians/s
\(\phi\) = 0
t = 8.50 s

So, we can find the value of amplitude by rearranging the formula:

x = A sin (\(\omega t + \phi\)) \(\rightarrow\) A = \(\frac{x}{sin (\omega t + \phi)}\)

A = \(\frac{x}{sin (\omega t + \phi)}\)

So, A = \(\frac{0.14 m}{sin [(\pi radians/s) (8.50s) + 0]}\)

A = \(\frac{0.140 m}{sin (8.50 \pi)}\)

Moreover, the sine of 8.50 \(\pi\) can be solved (by keeping in mind that the values is in radians) with a calculator:

Sin(8.50 \(\pi\)) = 1

So, the amplitude at time t is 8.50s is:

A = \(\frac{0.140 m}{sin(8.50 \pi)}\)

A = \(\frac{0.140 m}{1}\)

A = 0.140 m

Therefore, the amplitude of the pendulum’s oscillation is A =0.140 m = 14.0 cm.

Example 2

Assume that the head of a jack-in-the-box toy is bouncing upward and downward on a spring. Furthermore, the angular frequency of the oscillation is \(\omega\) = \(\pi /6 radians/s\), and the phase shift is \(\phi\) = 0 radians. Moreover, the amplitude of the bouncing is 5.00 cm. So, what is the position of the Jack-in-the-head, relative to the equilibrium position, at the following times?

a) 1.00 s

b) 6.00 s

Solution:

x = A sin (\(\omega t + \phi\))

x = (0.500 m) sin [(\(\pi /6 radians/s) (1.00 s)\) + 0]

x = (0.500 m) sin (\(\pi /6 radians/s \))

x = (0.500 m) (0.500)

x = 0.250 m
x = 2.50 cm

So, at time 1.00 s the head of the jack-in-the-box is 2.5 cm above the equilibrium position.

b) 6.00

x = A sin (\(\omega t + \phi\))

x = (0.500 m) sin [(\(\pi /6 radians/s) (6.00 s)\) + 0]

x = (0.500 m) sin (\(\pi /6 radians/s \))

x = (0.500 m) (0.00)

x = 0.00 m

So, at time t =6.00 s, the head of the-jack-in-the-box is at position 0.00 m that is the equilibrium position.

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Yashdeep tiwariKwame DavidumerMalek safrinRoger Carmichael Recent comment authors
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Roger Carmichael
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Typo Error>
Speed of Light, C = 299,792,458 m/s in vacuum
So U s/b C = 3 x 10^8 m/s
Not that C = 3 x 108 m/s
to imply C = 324 m/s
A bullet is faster than 324m/s

Malek safrin
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I have realy intrested to to this topic

umer
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umer

m=f/a correct this

B. Akshaya
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B. Akshaya

M=f/g

Kwame David
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Interesting studies

Yashdeep tiwari
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Yashdeep tiwari

It is already correct f= ma by second newton formula…

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