Question

The displacement x of a particle varies with time t as
$x=Asi{n}^2\omega t+Bco{s}^2\omega t+Csin\omega tcos\omega t$
For what values of A, B and C is the motion simple harmonic?

• A. All values of A, B and C with $C\ne 0$
• B. A = B, C = 2B
• C. A = – B, C = 2B
• D. A = B, C = 0

Answer 1

Answer: (A), (B), (C)

The correct options are
A

All values of A, B and C with $C\ne 0$

B

A = B, C = 2B

C

A = – B, C = 2B

The displacement equation can be rewritten as

$x=\frac{A}{2}(1-2coswt)+\frac{B}{2}(1+cos2wt)+\frac{C}{2}sin2wt$

or $x=\frac{1}{2}(A+B)+\frac{1}{2}(B-A)cos2wt+\frac{C}{2}sin2wt\dots \left(1\right)$

choice (a) : Equation (1) can be written as

$x=x_0+acos2wt+bsin2wt\dots \left(2\right)$

where $x_0=\frac{1}{2}(A+B),a=\frac{1}{2}(B-A)andb=\frac{C}{2}$

Equation (2) can be recasted as

$x=x_0+A_{0}sin(2wt+\phi )\dots \left(3\right)$

where $A_0=(a^2+b^2)\frac{1}{2}$ and $tan\phi =a/b$.

Equation (3) represents a simple harmonic motion of angular frequency 2ω,

amplitude = $x_0+A_0$ and phase constant $\phi .$

Choice (b): For A = B and C = 2B, Eq. (1) becomes

$x=B+Bsin2\omega t=B(1+sin2\omega t)$

This equation represents a simple harmonic motion of angular frequency 2ω.

Choice (c) : For A = – B and C = 2B, Eq. (1) becomes
$x=Bcos2\omega t+Bsin2\omega t$

which represents a simple harmonic motion of amplitude B, angular frequency 2ω.

Choice (d) : For A = B and C = 0, Eq. (1) reduces to
x = A.

which does not represent simple harmonic motion.
Hence the correct choices are (a), (b) and (c).