Function x = A sin^2(wt) + B cos^2(wt) + C sin(wt)cos(wt) represents SHM. Options: (a) For any value

Q: Why does the frequency of the SHM become $2\omega$ instead of $\omega$?

When we use double-angle identities to simplify $\sin^2 \omega t$ and $\cos^2 \omega t$, the resulting terms contain $\cos 2\omega t$ and $\sin 2\omega t$. Since the physical state repeats with this doubled frequency, the angular frequency of the resulting SHM is indeed $2\omega$.

Q: Can a function represent SHM if it has a constant offset like $x_0$?

Yes! A constant offset $x_0$ simply shifts the mean position of the oscillation away from the origin. The restoring force is still proportional to the displacement from this new mean position, i.e., $F \propto -(x - x_0)$, which is perfectly valid SHM.

Q: Why is option (c) not considered SHM?

If $A = B$ and $C = 0$, the equation simplifies to $x(t) = A(\sin^2 \omega t + \cos^2 \omega t) = A$. Since the displacement is constant and does not vary with time, there is no oscillation at all.

Q: How do we combine $A_1 \cos 2\omega t + A_2 \sin 2\omega t$ into a single harmonic term?

We can express it as $A_{\text{res}} \sin(2\omega t + \phi)$ where $A_{\text{res}} = \sqrt{A_1^2 + A_2^2}$ and $\tan \phi = A_1 / A_2$. This is a standard phasor addition technique for harmonic waves of the same frequency.

Q: What happens if the amplitude expression inside the square root becomes zero?

If $(B-A)^2 + C^2 = 0$, then both $B-A = 0$ (meaning $A=B$) and $C=0$. In this case, the amplitude is zero, and the system remains statically at its mean position without oscillating.

Question

Function $x = A \sin^2 \omega t + B \cos^2 \omega t + C \sin \omega t \cos \omega t$ represents SHM

Options:

(A) For any value of $A, B$ and $C$ (except $C = 0$)

(B) If $A = -B, C = 2B$, amplitude = $|B\sqrt{2}|$

(C) If $A = B; C = 0$

(D) If $A = B; C = 2B$, amplitude = $|B|$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

In this problem, we are presented with a mathematical function representing the displacement of a particle over time:

$$x(t) = A \sin^2 \omega t + B \cos^2 \omega t + C \sin \omega t \cos \omega t$$

Our objective is to determine the conditions under which this function represents **Simple Harmonic Motion (SHM)** and to find the corresponding amplitudes for those cases.

At first glance, the presence of squared terms like $\sin^2 \omega t$ and $\cos^2 \omega t$, along with the product term $\sin \omega t \cos \omega t$, might make the expression look non-linear and complex.

However, the key to solving such problems lies in transforming these quadratic trigonometric terms into linear terms using double-angle identities.

---

### The Master Equation & Simplification

Let us recall the standard double-angle trigonometric identities:

$$\sin^2 \omega t = \frac{1 - \cos 2\omega t}{2}$$

$$\cos^2 \omega t = \frac{1 + \cos 2\omega t}{2}$$

$$\sin \omega t \cos \omega t = \frac{\sin 2\omega t}{2}$$

Substituting these identities into our original displacement equation, we get:

$$x(t) = A \left( \frac{1 - \cos 2\omega t}{2} ight) + B \left( \frac{1 + \cos 2\omega t}{2} ight) + C \left( \frac{\sin 2\omega t}{2} ight)$$

Now, let us expand and group the terms systematically to separate the constant terms (which represent a shift in the mean position) from the time-varying harmonic terms:

$$x(t) = \frac{A}{2} - \frac{A}{2}\cos 2\omega t + \frac{B}{2} + \frac{B}{2}\cos 2\omega t + \frac{C}{2}\sin 2\omega t$$

$$x(t) = \left( \frac{A+B}{2} ight) + \left( \frac{B-A}{2} ight) \cos 2\omega t + \frac{C}{2} \sin 2\omega t$$

This is a remarkable result! The equation is now in the standard form of a shifted simple harmonic motion:

$$x(t) = x_0 + A_1 \cos 2\omega t + A_2 \sin 2\omega t$$

where:
- The **mean position** (offset) is $x_0 = \frac{A+B}{2}$
- The coefficient of the cosine term is $A_1 = \frac{B-A}{2}$
- The coefficient of the sine term is $A_2 = \frac{C}{2}$

Since both oscillating terms have the same angular frequency of $2\omega$, they combine to form a single harmonic oscillation. The resultant amplitude $A_{	ext{res}}$ of this combined motion is given by:

$$A_{	ext{res}} = \sqrt{A_1^2 + A_2^2}$$

$$A_{	ext{res}} = \sqrt{\left(\frac{B-A}{2}ight)^2 + \left(\frac{C}{2}ight)^2}$$

$$A_{	ext{res}} = \frac{1}{2}\sqrt{(B-A)^2 + C^2}$$

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### Evaluating the Options

Now that we have the general expressions for the mean position and the amplitude, let us evaluate each option one by one.

#### Option (a) & (c)
If we choose $A = B$ and $C = 0$ (as in option c), the displacement becomes:

$$x(t) = \frac{A+A}{2} + 0 + 0 = A$$

Since $x(t) = A$ is a constant value, there is no oscillation, and hence it does not represent SHM. This immediately rules out **Option (c)**.

Furthermore, because it fails for $A=B, C=0$, it cannot represent SHM for *any* arbitrary values of $A, B, C$. Thus, **Option (a)** is also incorrect.

#### Option (b)
Let us substitute the conditions $A = -B$ and $C = 2B$ into our general formulas:
- **Mean Position:** 
  $$x_0 = \frac{-B+B}{2} = 0$$
  This means the particle oscillates symmetrically about the origin ($x = 0$).
  
- **Resultant Amplitude:**
  $$A_{	ext{res}} = \frac{1}{2}\sqrt{(B - (-B))^2 + (2B)^2}$$
  $$A_{	ext{res}} = \frac{1}{2}\sqrt{(2B)^2 + 4B^2} = \frac{1}{2}\sqrt{8B^2} = |B\sqrt{2}|$$
  
This perfectly matches the statement in **Option (b)**. Therefore, **Option (b) is correct**.

#### Option (d)
Let us substitute the conditions $A = B$ and $C = 2B$ into our formulas:
- **Mean Position:**
  $$x_0 = \frac{B+B}{2} = B$$
  This means the particle oscillates about a shifted mean position of $x = B$.
  
- **Resultant Amplitude:**
  $$A_{	ext{res}} = \frac{1}{2}\sqrt{(B - B)^2 + (2B)^2}$$
  $$A_{	ext{res}} = \frac{1}{2}\sqrt{0 + 4B^2} = |B|$$
  
This perfectly matches the statement in **Option (d)**. Therefore, **Option (d) is correct**.

### Summary of Results
- The motion is simple harmonic with an angular frequency of $2\omega$.
- For $A = -B$ and $C = 2B$, the amplitude is $|B\sqrt{2}|$ about the mean position $x = 0$.
- For $A = B$ and $C = 2B$, the amplitude is $|B|$ about the mean position $x = B$.

Animated Solution for Physics - Oscillations: Function $x = A sin^{2} ω t + B cos^{2} ω t + C sin ω t cos ω t$ represents SHM

Select Answer:

Visualized Solution

Understanding the Objective

Recalling Trigonometric Identities

Substituting Identities into the Equation

Grouping and Simplifying Terms

Identifying the General SHM Form

Analyzing Option (a) and (c)

Analyzing Option (b)

Analyzing Option (d)

Final Conclusion

The Sigma Insight: Simple Harmonic Motion (SHM)

Analyzing the Setup

The Master Equation & Simplification

Evaluating the Options

# Option (a) & (c)

# Option (b)

# Option (d)

Summary of Results

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