Question

The amplitude of simple harmonic motion represented by the displacement equation $y\left( {cm} \right) = 4\left( {\sin 5\pi t + \sqrt 2 \cos 5\pi t} \right)$is:
A. 4 cm
B. $4\sqrt 2$cm
C. $4\sqrt 3$cm
D. $4\left( {\sqrt 2 + 1} \right)$cm

Answer 1

In this question, we need to determine the amplitude of simple harmonic motion represented by the displacement equation $y\left( {cm} \right) = 4\left( {\sin 5\pi t + \sqrt 2 \cos 5\pi t} \right)$. For this, we will compare the given expression with the standard equation of motion of the simple harmonic motion and then evaluate the resultant of the two amplitudes.

Complete step by step answer:
Given the equation of simple harmonic motion
$y\left( {cm} \right) = 4\left( {\sin 5\pi t + \sqrt 2 \cos 5\pi t} \right)$
Where y is the displacement in cm
We can write the given equation as
$y\left( {cm} \right) = 4\sin 5\pi t + 4\sqrt 2 \cos 5\pi t - - (i)$
Now we know that the general equation for the simple harmonic equation is given by the formula
$y = A\sin \left( {\omega t} \right) + B\cos \left( {\omega t} \right) - - (ii)$
Hence by comparing the equation (i) by equation (ii) we can write
$A = 4$
$B = 4\sqrt 2$
We also know that the amplitude for a simple harmonic equation is given as
$amplitude = \sqrt {{A^2} + {B^2}}$
Now substitute the values of A and B from the equation (i), so we get amplitude

$$amplitude = \sqrt {{{\left( 4 \right)}^2} + {{\left( {4\sqrt 2 } \right)}^2}} \\\ = {\kern 1pt} \sqrt {16 + 32} \\\ = \sqrt {48} \\\ = 4\sqrt 3 \\\$$

Hence the amplitude of simple harmonic motion $= 4\sqrt 3$

So, the correct answer is “Option C”.

Note:
A special type of periodic motion where the restoring force of the moving object is directly proportional to its displacement magnitude and which acts towards the objects equilibrium position is called simple harmonic motion. The general equation for a simple harmonic equation is given by $y = A\sin \left( {\omega t} \right) + B\cos \left( {\omega t} \right)$.