Question

The displacement y in centimeters is given in terms of time t in second by the equation: $y = 3\sin (3.14t) + 4\cos (3.14t)$ then the amplitude of SIMPLE HARMONIC MOTION is:

(A) 3cm
(B) 4cm
(C) 5cm
(D) 7cm

Answer 1

We know that the maximum value of $a\sin x + b\cos x$ is $\sqrt {{a^2} + {b^2}}$ . We will implicate the following formulae. The general equation of a Simple Harmonic Wave is in sin so the above will hold true.

Complete step-by-step solution

It is given to us that the equation of the SIMPLE HARMONIC MOTION is

$y = 3\sin (3.14t) + 4\cos (3.14t)$ .
Here $a = 3$ and $b = 4$

Using $\sqrt {{a^2} + {b^2}}$
$\Rightarrow Amplitude = \sqrt {{3^2} + {4^2}} = \sqrt {25} = 5$
The maximum value of the given function is 5 and the maximum displacement in SIMPLE HARMONIC MOTION is the Amplitude of that SIMPLE HARMONIC MOTION.

Therefore, the answer is option C, 5cm.

Note The theta given inside the cos and the sin functions is equal therefore we are able to apply $\sqrt {{a^2} + {b^2}}$ . If this were not the case then convert sin function to cos function or vice-versa and then add the corresponding functions to obtain the required value of amplitude.

Additional Information Simple Harmonic Motions is a type of periodic motion in which the acceleration of the body is directly proportional to its displacement from the mean position.
With the frequency of oscillations being the constant.
$a = {\omega ^2}x$
And its general equation is given by $x = A\sin (\omega t + \phi )$