Question

A particle is executing SHM with the equation \[x=3\cos \omega t+4\sin \omega t\]. Find the phase shift and the amplitude.
\[\begin{align}
  & \text{A) }{{50}^{0}},5\text{ units} \\
 & \text{B) }{{37}^{0}},\text{3}\text{.5 units} \\
 & \text{C) }{{53}^{0}},\text{3}\text{.5 units} \\
 & \text{D) 3}{{\text{7}}^{0}},\text{5 units} \\
\end{align}\]

Answer 1

Hint: A particle undergoing simple harmonic motion has an amplitude and phase shift which is dependent on the time. Here, we have the solution to the equation of the particle motion in SHM. We can find each of the physical quantities from the equation of motion directly.

Complete step-by-step solution
The solution of the equation of motion of a particle undergoing simple harmonic motion consists of the physical quantities such as its amplitude, frequency, phase change, and Time period. Properly substituting the values, we can find the measure of these quantities at any instant in time. The position of the particle solved from the equation of a particle in SHM is as follows:
\[x=a\sin (\omega t+\phi )\]
Where x is the position of the particle at any instant ‘t’,
a is the maximum displacement of the particle from mean position or amplitude,
\[\omega \] is the angular frequency,
\[\phi \] is the phase difference.
Now let us consider the position of the particle given in this situation. It is given as –
\[x=3\cos \omega t+4\sin \omega t\]
We can understand from the equation that the particle undergoes a motion which is periodic with respect to both the sine and cosine functions.
Let us substitute some values to the terms involved for the ease of calculation. For instance, let
\[\begin{align}
  & 3={{x}_{0}}\sin \phi \text{ --(1)} \\
 & \text{and } \\
 & 4={{x}_{0}}\cos \phi \text{ --(2)} \\
\end{align}\]
The solution of the equation of motion thus, becomes,
\[\begin{align}
  & x={{x}_{0}}\sin \phi \cos \omega t+{{x}_{0}}\cos \phi \sin \omega t \\
 & \text{but, } \\
 & \sin A\cos B+\cos AsinB=\sin (A+B) \\
 & \Rightarrow \text{ }x={{x}_{0}}\sin (\omega t+\phi ) \\
\end{align}\]
So, we have reached a simpler version of the solution of the equation of motion which can be easily compared to the standard equation.
We can find the maximum amplitude involved by squaring and adding (1) and (2) as
\[\begin{align}
  & {{x}_{0}}^{2}{{\sin }^{2}}\omega t+{{x}_{0}}^{2}{{\cos }^{2}}\omega t=9+16=25 \\
 & \Rightarrow \text{ }{{x}_{0}}^{2}({{\sin }^{2}}\omega t+{{\cos }^{2}}\omega t)=25 \\
 & \Rightarrow \text{ }{{x}_{0}}=5 \\
\end{align}\]
Also, we can find the phase shift by dividing (1) by (2) –
\[\begin{align}
  & \dfrac{{{x}_{0}}\sin \phi }{{{x}_{0}}\cos \phi }=\dfrac{3}{4} \\
 & \Rightarrow \text{ }\tan \phi =\dfrac{3}{4} \\
 & \Rightarrow \text{ }\phi ={{\tan }^{-1}}\dfrac{3}{4} \\
 & \Rightarrow \text{ }\phi ={{37}^{0}} \\
\end{align}\]
Now, we get the solution of the equation of the particle motion in SHM as –
\[x=5\sin (\omega t+{{37}^{0}})\]
The required amplitude is 5 units and the phase shift is \[{{37}^{0}}\].
The correct is option D.

Additional Information: In nature, there exists no perfect SHM due to the presence of drag and frictional forces.

Note: We can find the frequency, time period, velocity, and acceleration of the system by substituting proper values and relations in the solution of the equation of the motion. Also, the equation of motion is also solved using the exponential function.