Amplitude is the maximum displacement of a particle or wave from its equilibrium (mean) position. It represents the greatest distance an object moves away from its central position during periodic motion.
- SI unit of amplitude is the meter (m)
- Amplitude is always taken as a positive quantity
- It indicates the energy of oscillation (larger amplitude → more energy)
- The total distance between the two extreme positions is 2A.
- Amplitude is a key parameter in oscillations and wave motion, determining the strength and intensity of the motion
For example, in a pendulum, the bob swings to a maximum distance on either side of the mean position. This maximum distance is called the amplitude and is denoted by A
Mathematical Representation
Periodic motions such as waves and spring–mass systems are described using sine or cosine functions:
\boxed {x = A \sin(\omega t + \phi)} or\boxed {x = A \cos(\omega t + \phi)}
where,
- x = displacement of wave (meter)
- A = amplitude
- ω = angular frequency (rad/s)
- t = time period
- ϕ = phase angle
Solved 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
x = 0.140 m
ω = π radians/s
ϕ = 0
t = 8.50 s
the value of amplitude by rearranging the formula
x = A \sin(\omega t + \phi)
A = \frac{x}{\sin(\omega t + \phi)} Substituting the given values
A = \frac{0.140}{\sin[(\pi)(8.50) + 0]}
A = \frac{0.140}{\sin(8.50\pi)}
\sin(8.50\pi) = 1
A = \frac{0.140}{1}
A = 0.140\,\text{m}
\boxed{A = 0.140\,\text{m} = 14.0\,\text{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:
x = A \sin(\omega t + \phi)
x = (0.050)\sin\left[\left(\frac{\pi}{6}\right)(6.00) + 0\right]
x = (0.050)\sin(\pi)
\sin(\pi) = 0
\boxed{x = 0.00\,\text{m}}
Problem 3: If y = 6 cos(7t + 1), then it is a wave. Find its amplitude.
Solution: Equation of wave y = 6cos(7t + 1)
amplitude formula,
x= A cos (ωt + ϕ)
comparing it with the wave equation:
A = 6
ω = 7
ϕ = 1
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)
the formula for amplitude,
x = A sin(ωt + ϕ)
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/6 radians/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?
Solution:
x = A \sin(\omega t + \phi)
x = (0.05)\sin\left[\left(\frac{\pi}{6}\right)(1.00) + 0\right]
x = (0.05)\sin\left(\frac{\pi}{6}\right)
\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}
x = (0.05)\times 0.5
x = 0.025\,\text{m}
\boxed{x = 2.5\,\text{cm}}
Unsolved problems
Problem 1: A pendulum swings with a maximum angular displacement of 15 degrees. What is the amplitude of the pendulum in meters if the length of the pendulum is 2 meters?
Problem 2: A sound wave in air has a pressure amplitude of 0.01 Pa. If the wave's frequency is 1000 Hz, what is its amplitude in terms of displacement?
Problem 3: A wave on a string has a wavelength of 2 meters and a frequency of 5 Hz. If the maximum displacement of a point on the string is 0.1 meters, what is the amplitude of the wave?
Problem 4: The intensity of a light wave is directly proportional to the square of its amplitude. If the intensity is increased by a factor of 4, by what factor does the amplitude increase?