Question

A particle in SHM is described by the displacement function $ x\left( t \right){\text{ }} = {\text{ }}acos\left( {ax + {\text{ }}\theta } \right) $ . if the initial position of the particle is 1cm and its initial velocity is $ \pi cm/s $ . the angular frequency of the particles is $ \pi {\text{ }}rad/s $ , then its amplitude is ?
(A) $ 1cm $
(B) $ \sqrt 2 cm $
(C) $ 2cm $
(D) $ 2.5cm $

Answer 1

Hint: The function for displacement is given in the question, we can get the equation for velocity by differentiating the displacement with time. The value of displacement and velocity of the particle at time zero is also given, this value can be substituted in the steps while deriving the answer. After finding all the parameters the amplitude can be found out.

Complete answer:
From the question displacement function of the particle is given by , $ x\left( t \right){\text{ }} = {\text{ }}acos\left( {ax + {\text{ }}\theta } \right) $
At time $ t = 0 $ , position $ x(0) = 1cm $ and velocity $ v(0) = \pi cm/s $
To get velocity, we differentiate position with respect to time
 $ v(t) = - a(\omega )sin(\omega t + \varphi ) - - - - - (1) $
At time $ t = 0 $
 $ x(0) = a\cos (\phi ) - - - - - (2) $
 $ v(0) = - a\omega sin(\phi ) - - - - - (3) $
Dividing (3) by (2) we get
 $ \dfrac{{v(0)}}{{x(0)}} = - \omega \tan \phi - - - - - (4) $
But we know that $ x(0) = 1cm $ and $ v(0) = \pi cm/s $
Therefore , (4) becomes , $ \pi = - \omega \tan \phi - - - - - (5) $
Given angular frequency $ \left( \omega \right) $ is $ \pi {\text{ }}rad/s $
(5) becomes , $ \tan \phi = - 1 $
 $ \Rightarrow \phi = \dfrac{{3\pi }}{4} $ or $ \dfrac{{7\pi }}{4} $
But at $ t = 0 $ , $ x(0) = A\cos (\phi ) = 1 $
 $ \Rightarrow \cos (\phi ) $ is positive
 $ \Rightarrow \phi = \dfrac{{7\pi }}{4} $
 $ \Rightarrow a\cos \left( {\dfrac{{7\pi }}{4}} \right) = 1 $
 $ \Rightarrow a = \dfrac{1}{{\left( {{{\cos }^{ - 1}}\left( {\dfrac{{7\pi }}{4}} \right)} \right)}} $
 $ \Rightarrow a = \dfrac{1}{{0.707}} $
 $ \Rightarrow a = 1.414 = \sqrt 2 $ , this is the amplitude.
The answer is option (B), $ \sqrt 2 $ .

Note:
SHM stands for Simple Harmonic Motion, which is described as a motion in which the restoring force is proportional to the body's displacement from its mean position. This restorative force always moves in the direction of the mean position. The period T is the amount of time it takes the particle to complete one oscillation and return to its initial location. The angular frequency is denoted by $ \omega = \dfrac{{2\pi }}{T} $ , where T is the period. Radians per second are used to measure angular frequency.