Question

The motion of a particle is given by x = A.sin⍵t+B.cos⍵t. The motion of the particle is
(a) not simple harmonic
(b) simple harmonic with amplitude A+B
(c) simple harmonic with amplitude (A+B)/2
(d) simple harmonic with amplitude √(A²+B²).

Answer 1

Given Equation,

 x = A.sin⍵t+B.cos⍵t.

Let ωt = α

Now, multiplying and dividing by $\sqrt{A^2 - B^2}$

x = $\sqrt{A^2 - B^2} [\frac{A}{\sqrt{A^2 - B^2}}Sin\alpha  + \frac{B}{\sqrt{A^2 - B^2}}Cos\alpha ]$

Let, $\frac{A}{\sqrt{A^2 - B^2}} = Cos\beta$

and $\frac{B}{\sqrt{A^2 - B^2}} = Sin\beta$

Therefore,  

x = √(A²+B²){sinαα.cosβ+cosα.sinβ}

x = √(A²+B²){Sin(α + β)}

x = √(A²+B²){Sin(ωt + β)}

Now, Comparing it with x = A Sin(ωt + Φ),

A = √(A²+B²), therefore, Amplitude = √(A²+B²)

Hence, Option (d) is correct.

Hope it helps.

Answer 2

Answer:

Refer to the above attachments.

Thank you.