Amplitude Formula

The largest deviation of a variable from its mean value is referred to as amplitude. It is the largest displacement from a particle’s mean location in to and fro motion around a mean position. Periodic pressure variations, periodic current or voltage variations, periodic variations in electric or magnetic fields, and so on all have amplitudes.

Amplitude does not have a specific formula. It can be obtained by equations or graphical representations of such variations.

What is an Amplitude?

The highest displacement of the waves is referred to as amplitude. In addition, you will learn about amplitude, amplitude formula, formula derivation, and a solved example in this course. Furthermore, you will be able to comprehend amplitude after completing the topic.

Amplitude refers to the greatest deviation from equilibrium that an item in periodic motion might display. A pendulum, for example, swings past its equilibrium point (straight down) before reaching its maximum distance from the centre.

Furthermore, the amplitude’s distance is A. Furthermore, the pendulum’s complete range has a magnitude of 2A. Waves and springs, for example, follow a periodic motion. Furthermore, because the sine function oscillates between +1 and -1, it may be used to depict periodic motion.

SI Unit: The metre is the most notable amplitude unit (m).

Formula for Amplitude

The amplitude of a variable is the biggest variation from its mean value. The amplitude formula can be used to calculate the sine and cosine functions. Amplitude is represented by the letter A. The sine (or cosine) function has the following formula:

x = A sin (ωt + ϕ) 

or

x = A cos (ωt + ϕ)

where,

• x = displacement of wave (meter)
• A = amplitude
• ω = angular frequency (rad/s)
• t = time period
• ϕ = phase angle

The amplitude formula is also known as the average of the maximum and minimum values of a sine or cosine function. The absolute amplitude value is always used.

Sample Problems

Problem 1: Consider a pendulum that swings back and forth. In addition, the phase shift is 0 radians. Furthermore, the pendulum is 14.0 cm or x = 0.140 m, and the time is t = 8.50 s. So, what is the oscillation’s amplitude?

Solution:

Given that,

x = 0.140 m

ω = π radians/s

ϕ = 0

t = 8.50 s

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

x = A sin (ωt+ϕ) → A = xsin(ωt+ϕ)

A = xsin(ωt+ϕ)

So, A = 0.14msin[(πradians/s)(8.50s)+0]

A = 0.140msin(8.50π)

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

Sin(8.50 π) = 1

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

A = 0.140msin(8.50π)

A = 0.140m1

A = 0.140 m

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

Problem 2: Assume a spring is bouncing the head of a jack-in-the-box toy upward and downward. In addition, the oscillation’s angular frequency is π/6 radians/s, with a phase shift (ϕ) of 0 radians. The bouncing also has a 5.00 cm amplitude. Where does the Jack-in-the-head stand in relation to the equilibrium position in 6 s?

Solution:

Since, as we know that:

x = A sin (ωt+ϕ)

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

x = (0.500 m) sin (π/6radians/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.

Problem 3: If y = 6 cos (7t + 1) is a wave. Find its amplitude.

Solution:

Given: equation of wave y = 6cos(7t + 1)

Using amplitude formula,

x= A cos (ωt + ϕ)

On comparing it with the wave equation:

A = 6

ω = 7

ϕ = 1

Therefore, the amplitude of the wave = 6 units. 

Problem 4: A wave is y = 2sin(4t). Find out its amplitude.

Solution:

The wave equation y = 2sin(4t)

Using the formula for amplitude,

x = A sin(ωt + ϕ)

When comparing the wave equation to the equation of motion,

A = 2

ω = 4

ϕ = 0

As a result, the amplitude of the wave is 2 units.

Problem 5: Consider a jack-in-the-box toy with its head bouncing up and down on a spring. Furthermore, the oscillation’s angular frequency is = π/6radians/s, and the phase shift is ϕ= 0 radians. Furthermore, the bouncing has a 5.00 cm amplitude. So, where does the Jack-in-the-head stand in relation to the equilibrium position in 1s.

Answer:

x = A sin (ωt+ϕ)

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

x = (0.500 m) sin (π/6radians/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.