How To Find The Amplitude Of A Wave: Formula, Problems, Examples And Facts 


In this post we will analyze different aspects of wave and how to find the amplitude of a wave.  

The highest elevation from the equilibrium point reached by a wave is described as amplitude. The letter A is used to express it. Amplitude may be expressed as y=Asinωt+Φ 

Here,  A is amplitude of wave 

           y is displacement of wave 

           ω is angular frequency expressed as ω=2π/t

           π is phase difference  

The peak point of a wave is the maximal amplitude of vertical displacement throughout a cycle. On a surface wave, a peak is a point when the medium’s displacement is greatest 

The trough point of a wave represents the greatest downward displacement throughout a cycle. On a wavefront, a trough is a point where the medium’s displacement is greatest in the downward direction. 

The amount of power utilized to begin the waves determines the amplitude. Larger amplitude waves have more strength and intensity. 

A wave’s amplitude and frequency are proportional, with the amplitude being proportional to the frequency.  When the frequency rises, the amplitude falls. When the frequency lowers, the amplitude rises. 

The speed with which a pulse passes is unaffected by its amplitude. Waves A and B both travel at the same speed. The velocity of a wave is wholly determined by changes in the properties of the medium it travels through. 

Amplitude of a wave formula 

Formula to find amplitude of wave is Position = amplitude * sine function (angular frequency * time + phase difference)

 Amplitude of a wave is found directly from mathematical form of wave that is y=Asin(ωt +Φ ). Amplitude is equal to A.  

The above equation is formula to find amplitude of a wave. This formula can be used to find time period T, frequency ω, displacement y, phase difference Φ and wavelength λ  of a wave. 

How to find the amplitude of a wave on a graph? 

By calculating the distance between crest and equilibrium or trough and equilibrium from a wave graphic. 

In the graphical approach, the length from equilibrium to trough equals amplitude or the length from equilibrium to crest, so we can simply measure the distance from the plot to get the amplitude of the wave. 

How to find the amplitude of a longitudinal wave?  

The greatest displacement by wave from the equilibrium point is measured. 

The largest displacement of a component from its resting point affects the amplitude of a longitudinal wave, such as a sound pulse. A wave is considered to be wetted when its amplitude steadily drops as its power is dissipated. 

Because determining the height at which particles travel is difficult, the amplitude is usually expressed in terms of transverse waves. A longitudinal wave’s amplitude is parallel to the plane of motion of the wave. A longitudinal wave is one that has a periodic disturbance or oscillation in the same plane as the wave’s travel.

A compression wave traverses its size, preceded by an elongation when a circular spring is squeezed on one side and then released on the other; any location on any loop of the spring will stream with the wave and back along the same route, passing through the neutral state and then undoing its movement.

The gas along the line of flow of the sound wave is compressed and rarefied as the sound wave swings back and forth. The longitudinal character of P (primary) seismic waves is identical to that of P (secondary) seismic waves.

Apart from a gradual change in phase (q.v.) of vibration—that is, every particle finishes its cycle of response at a later time—every particle of matter oscillates concerning its normal resting position, and other than the axial direction of transmission in a longitudinal wave, and all particles involved in the wave motion act precisely in the same way.

In the axis of transmission, the combined movements create alternating zones of compression and rarefaction to proceed. 

How to find the amplitude of a transverse wave? 

Measuring the maximum displacement by wave from equilibrium point. 

The space between the equilibrium point and either the apex (high point of the wave) or the lower side (low point of the wave) is the amplitude of a transverse wave (low point of the wave.) 

 A transverse wave’s amplitude is perpendicular to the wave’s plane of motion. 

The amplitude of a wave created by a more severe disturbance is greater. Consider tossing a little stone into a peaceful pond. Little waves will develop from the disruption in circular rings. The waves have a very little amplitude and energy. Toss a heavy pebble into the water as an experiment. Extremely large waves will result from the disturbance. These are waves with a bunch of power and a lot of amplitude. 

The amplitude of a wave is proportional to the quantity of power it contains. A large-amplitude wave carries a lot of energy, whereas a short-amplitude pulse only carries a little amount. The mean quantity of power going across a unit area per unit of time in one axis is the intensity of a pulse.

The intensity of the noise grows as the amplitude of the sound wave rises. Sounds with higher intensity are considered louder. Sound intensity comparisons are usually stated in decibels (dB). 

Problems 

Problem 1 

Consider a clock swinging backward and forward. The oscillation’s angular frequency is equal to 2π radians/s, and the phase difference is zero radians. Furthermore, the duration is t = 10 s, and the length of the clock is 12.0 cm or x = 0.120 m. Therefore, what will be the amplitude of oscillations? 

Solution: 

First, we will write all the given values and then will find out the amplitude by putting the values in amplitude formula. 

Given:  

Angular frequency, ω = 2π (in radians/s) 

Phase difference, Φ=0 (in radians) 

Time period, t=10 (in seconds) 

y=0.120 m

Now we will put the above values in amplitude formula. 

y=A sin(ω t +π )

0.120 (m)= A sin(2π(radians/second) * t(seconds) + 0 )

A= 0.120(m)/Sin(2π(radians/seconds) (10 seconds)+0 )

A=0.120(m)/Sin 20π

We can calculate Sin(20π) directly from calculator and it is equal to sin(20π)= 0.88965. 

Now further solving,  

A=0.120(m)/0.88965

A= 0.1345 m

So, amplitude of wave is 0.1345 meters or 13.45cm. 

Problem 2 

Consider a clock swinging backward and forward. The oscillation’s angular frequency is equal to 2π radians/s, and the phase difference is zero radians. Furthermore, the duration is t = 15 s, and the length of the clock is 15.0 cm or x = 0.150 m. Therefore, what will be the amplitude of oscillations? 

Solution: 

First, we will write all the given values and then will find out the amplitude by putting the values in amplitude formula. 

Given:  

Angular frequency, ω = 2π (in radians/s) 

Phase difference, Φ=0 (in radians) 

Time period, t=15 seconds (in seconds) 

y=0.150 m 

Now we will put the above values in amplitude formula. 

y=Asin(ω t + π )

0.150 (m)= A sin(2π(radians/second) t(seconds) + 0 )

A= 0.150(m)/Sin(2π(radians/seconds) (15 seconds)+0 )

A=0.150(m)/Sin(30π)

We can calculate Sin 30π directly from calculator and it is equal to Sin 30π= 0.99725.

Now further solving,  

A= 0.150(m)/0.99725

A= 0.1503 m

So, amplitude of wave is 0.1503 meters or 15.03cm. 

Problem 3 

Consider a pendulum swinging backward and forward. The oscillation’s angular frequency is equal to 2π radians/s, and the phase difference is zero radians. Furthermore, the duration is t = 10 s, and the length of the clock is 30.0 cm or x = 0.300 m. Find out the amplitude of the wave.

Solution: 

First, we will write all the given values and then will find out the amplitude by putting the values in amplitude formula. 

Given:  

Angular frequency, ω= 2π(in radians/s) 

Phase difference, π=0 (in radians) 

Time period, t=5 seconds 

y=0.300 m

Now we will put the above values in amplitude formula. 

y=A sin(ω t + π)

0.300 (m)= A sin(2π (radians/second)* t(seconds) + 0)

A= {0.300(m)/Sin(2π(radians/seconds)(5 seconds)+0 )

A= {0.300(m)/Sin(10π)}

We can calculate Sin(10π) directly from calculator and it is equal to Sin(10π)= 0.52123.

Now further solving,  

A= 0.300(m)/0.52123

A= 0.5755 m

So, amplitude of wave is 0.5755 meters or 57.55 cm. 

Frequently asked questions |FAQs 

Ques. What is the relationship between amplitude and wavelength? 

Ans. A wave’s wavelength and amplitude are two critical parameters.

The elevation of a pulse is defined as the distance between the apex (or crest) and the bottom spot on the wave (trough). The space between one peak and the following is the wavelength of a wave. 

A waveform’s frequency is equivalent to its wavelength. Frequency is measured in hertz (Hz), or vibrations per second, and refers to the number of waves that pass-through a given area in a given amount of time. The frequencies of smaller wavelengths are higher, whereas the frequencies of larger wavelengths are lower. 

Ques. How is the amplitude of a wave affected by its energy? 

Ans. The amplitude squared is equivalent to the energy provided by a pulse.

Whatever amplitude fluctuation occurs, the power is affected by the square of that effect. This means that quadrupling the power equals doubling the amplitude. 

The amplitude of a huge power wave is large, whereas the amplitude of a shorter energy wave is low. The highest level of disruption of a component on the medium from its equilibrium position is the amplitude of a wave. The reasoning for the energy-amplitude link is as follows: The first loop obtains a preliminary amount of displacement when a slinky is stretched horizontally and a transverse wave is introduced into it. 

The force applied to the ring by the individual in effort to shift it to a particular distance from the rest causes the dislocation. The more energy a person puts into the wave, the more labor he or she puts into the initial loop. The greater the displacement given to the first loop, the more work done on it.

The amplitude of the first coil is proportional to the amount of dislocation given to it. Finally, the amplitude of a transverse wave is equal to the amount of energy carried across the medium by the pulse. A transverse wave’s wavelength, frequency, or speed are unaffected by putting a lot of power into it. 

SAKSHI KM

I am Sakshi Sharma, I have completed my post-graduation in applied physics. I like to explore in different areas and article writing is one of them. In my articles, I try to present physics in most understanding manner for the readers.

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