Question

The equation of an SHM is given by $ x = 3\sin 5\pi t + 4\cos 5\pi t $ , where $ x $ is in cm and time $ t $ is in seconds. Find the phase constant of the motion.
(A) $ 45^\circ $
(B) $ 30^\circ $
(C) $ 53.1^\circ $
(D) $ 60^\circ $

Answer 1

Hint :Phase constant shows the displacement of the initial position of the body from the origin or equilibrium position. Thus by comparing the given equation with the standard equation of SHM we can obtain the value of the phase constant.

Complete Step By Step Answer:
The general equation of SHM is given as
 $ x = A\sin (\omega t + \phi ) $
Where $ A $ is the amplitude or maximum height attained during the motion
 $ \omega $ is the angular frequency of the oscillations
 $ t $ is the time
 $ \phi $ is the phase constant of the motion.
The phase constant of the motion can be defined as the distance of the initial position of the object from the origin or the equilibrium position.
Thus if the object is in an equilibrium position when it starts the motion, the phase constant is said to be zero.
Here, from the general equation of SHM, let us expand the $ \sin $ function,
 $ \therefore x = A[\sin \omega t\cos \phi + \cos \omega t\sin \phi ] $
 $ \therefore x = A\sin \omega t\cos \phi + A\cos \omega t\sin \phi $
With the obtained equation, we can compare the equation with the given equation as both the equations have time as a variable in the sine and cosine functions.
The comparison can be shown as,
 $ A\sin \omega t\cos \phi = 3\sin 5\pi t $ and $ A\cos \omega t\sin \phi = 4\cos 5\pi t $
From the above two equations, the angular frequency can be found to be $ \omega = 5\pi $
Now, consider the first comparison equation,
 $ \therefore A\cos \phi \sin 5\pi t = 3\sin 5\pi t $
 $ \therefore A\cos \phi = 3 $ …… $ (1) $
Similarly, consider the second comparison equation
 $ \therefore A\sin \phi \cos 5\pi t = 4\cos 5\pi t $
 $ \therefore A\sin \phi = 4 $ …… $ (2) $
Dividing the equation $ (2) $ by equation $ (1) $ ,
 $ \therefore \dfrac{{A\sin \phi }}{{A\cos \phi }} = \dfrac{4}{3} $
 $ \therefore \tan \phi = \dfrac{4}{3} $
As the value is greater than $ 1 $ , we can understand that the phase constant is greater than $ 45^\circ $
Also, we know that,
 $ \dfrac{4}{3} = 1.333 < 1.73 $
But, we know $ 1.73 = \sqrt 3 $
Thus, we can write the inequality as
 $ \dfrac{4}{3} < \sqrt 3 $
Now, we know from the standard values of the trigonometric tables that $ \sqrt 3 = \tan 60^\circ $
Thus, we understand from this that the value of phase constant lies between $ 45^\circ $ and $ 60^\circ $
Hence, the correct answer is Option $ (C) $ .

Note :
The value of the phase constant can be directly calculated from the equation $ \tan \phi = \dfrac{4}{3} $ by the use of a scientific calculator, but as we don’t have the facility available, we will calculate the angle by the inequalities.