Question

A simple wave motion is represented by $y(t) = 5\left( {sin(4\pi t) + \sqrt 3 cos4\pi t} \right)$. Its amplitude is:
A. 5
B. $5\sqrt 3 $
C. $10\sqrt 3 $
D. 10

Answer 1

Hint: Try converting the equation into standard form of SHM: $y(x,t) = Asin(wt + \phi )$. Also recollect the trigonometric identity: $sin(A)cos(B) + cos(A)sin(B) = sin(A + B)$.

Complete step by step answer:
The wave equation of a simple harmonic motion contains all the information regarding the state of the wave. We can find the amplitude, velocity, phase, frequency, wavelength and even the position and momentum of the particle.
The amplitude of an SHM is the maximum displacement of a particle from its mean position.
If $y(t) = Asin(wt + \phi )$ is the wave equation, we see that the value of $y$ becomes maximum when $sin(wt + \phi )$ becomes maximum. Since the maximum value of
$sin$ is 1, we see the amplitude of the wave is A.
Here in the question, the SHM equation is given as :
$y(t) = 5\left( {sin(4\pi t) + \sqrt 3 cos4\pi t} \right)$
We can divide and multiply this equation by 2 to get
$y(t) = 10\left( {\dfrac{1}{2}sin(4\pi t) + \dfrac{{\sqrt 3 }}{2}cos4\pi t} \right)$
we know $cos\left( {\dfrac{\pi }{3}} \right) = \dfrac{1}{2}$ and $cos\left( {\dfrac{\pi }{3}} \right) = \dfrac{{\sqrt 3 }}{2}$
So let's replace these values in eqn (1)

$y(t) = 10\left( {\dfrac{1}{2}sin\left( {4\pi t} \right) + \dfrac{{\sqrt 3 }}{2}cos\left( {4\pi t} \right)} \right)$

Since $sin(A)cos(B) + cos(A)sin(B) = sin(A + B)$,

we can write the above equation as:
$y(t) = 10sin\left( {4\pi t + \dfrac{{\pi}}{3}} \right)$

Now, we know $sin\left( {4\pi t + \dfrac{\pi }{3}} \right)$ has a maximum value of 1.
So the maximum possible displacement and hence, the amplitude is 10.

Additional Information:
A simple harmonic motion has a fixed energy associated with it. At all points of its oscillation, the particle has a total energy of $\dfrac{1}{2}A{\omega ^2}$
If a particle of mass m is oscillating under a Force \[F = - kx\], Then energy conservation gives
$\dfrac{1}{2}m{v^2} + \dfrac{1}{2}k{x^2} = \dfrac{1}{2}A{\omega ^2}$ where $v$ is its velocity at any instant and $x$ its displacement from mean position.

Note: Here we were able to quickly convert just by looking at the equation. In general, it need not be this easy and we may have to use derivatives to find the point of maxima. The displacement from mean position to this point of maxima gives the amplitude.