The x-t graph of a particle undergoing simple harmonic motion is shown below. The acceleration of th

Q: Why did we choose the sine function instead of the cosine function for the displacement equation?

We chose the sine function because the graph starts at $x = 0$ at $t = 0$ and goes upwards in the positive direction. A cosine function would start at the maximum positive displacement ($x = A$) at $t = 0$.

Q: What does the negative sign in the final acceleration value represent?

The negative sign represents the restoring nature of SHM. It indicates that the acceleration vector is directed opposite to the displacement vector, always pointing back towards the mean position.

Q: Can we solve this problem using a phasor diagram?

Yes! On a phasor circle of radius $A = 1\text{ cm}$, the phasor rotates at $\omega = \pi/4\text{ rad/s}$. At $t = 4/3\text{ s}$, the phase angle is $\theta = \omega t = 60^\circ$. The projection on the vertical axis gives the displacement $x = A \sin 60^\circ = \sqrt{3}/2\text{ cm}$. Then, using $a = -\omega^2 x$, we get the same result.

Q: Why is the unit of acceleration in $\text{cm s}^{-2}$ instead of $\text{m s}^{-2}$?

The displacement on the vertical axis of the graph is given in centimeters (cm). Therefore, the amplitude is in centimeters, which makes the resulting acceleration unit centimeters per second squared ($\text{cm s}^{-2}$).

Q: What would be the acceleration of the particle at $t = 4\text{ s}$?

At $t = 4\text{ s}$, the particle is at the mean position ($x = 0$) as seen from the graph. Since the displacement is zero, the restoring force and acceleration are also zero ($a = -\omega^2 (0) = 0$).

Question

The $x-t$ graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at $t = 4/3	ext{ s}$ is

Options:

(A) $\frac{\sqrt{3}}{32} \pi^2 	ext{ cm s}^{-2}$

(B) $-\frac{\pi^2}{32} 	ext{ cm s}^{-2}$

(C) $\frac{\pi^2}{32} 	ext{ cm s}^{-2}$

(D) $-\frac{\sqrt{3}}{32} \pi^2 	ext{ cm s}^{-2}$

Sigma Percentile · Accepted Answer

### Introduction to Simple Harmonic Motion (SHM) Graphs

Simple Harmonic Motion (SHM) is one of the most fundamental and elegant concepts in physics.

It describes the back-and-forth oscillation of a particle about a stable equilibrium position under the influence of a restoring force that is directly proportional to the displacement.

When we plot this motion over time, we get a beautiful sinusoidal wave.

In this problem, we are given a displacement-time ($x-t$) graph of a particle executing SHM, and we are tasked with finding its acceleration at a specific instant, $t = 4/3	ext{ s}$.

Let's embark on this journey to decode the graph and extract the physics hidden within it.

---

### Analyzing the Setup

Let's look closely at the given $x-t$ graph.

At $t = 0$, the displacement $x$ is $0$, and the curve immediately rises into the positive region.

This tells us that the particle starts its journey from the **mean position** and moves in the positive direction.

The maximum height of the wave represents the **amplitude** ($A$) of the oscillation.

From the vertical axis, we can clearly see that:

$$A = 1	ext{ cm}$$

Next, let's find the **time period** ($T$), which is the time taken to complete one full cycle of oscillation.

Following the wave from $t = 0$, it goes up to a peak, comes down through zero to a trough, and returns to zero, completing one full cycle at:

$$T = 8	ext{ s}$$

With these two parameters in hand, we have unlocked the DNA of this simple harmonic oscillator.

---

### The Master Equation

To find the acceleration at any time $t$, we first need to write down the mathematical equation describing the displacement $x(t)$.

Since the particle starts at the mean position ($x = 0$) at $t = 0$ and moves in the positive direction, its displacement is best described by a sine function:

$$x(t) = A \sin(\omega t)$$

Here, $\omega$ is the **angular frequency**, which represents how fast the phase of the oscillation changes.

We can calculate $\omega$ using the relation:

$$\omega = \frac{2\pi}{T}$$

Substituting $T = 8	ext{ s}$ into this formula:

$$\omega = \frac{2\pi}{8} = \frac{\pi}{4}	ext{ rad s}^{-1}$$

Now, substituting $A = 1	ext{ cm}$ and $\omega = \frac{\pi}{4}	ext{ rad s}^{-1}$ back into our displacement equation, we get:

$$x(t) = \sin\left(\frac{\pi}{4} tight)$$

---

### Finding the Acceleration

In simple harmonic motion, acceleration ($a$) is always directly proportional to displacement ($x$) but acts in the opposite direction.

This fundamental relationship is given by the equation:

$$a(t) = -\omega^2 x(t)$$

Alternatively, we can derive this by differentiating the displacement equation twice with respect to time.

Let's do that to see the beautiful connection between calculus and physics:

1. **Velocity ($v$)** is the first derivative of displacement:

$$v(t) = \frac{dx}{dt} = \frac{\pi}{4} \cos\left(\frac{\pi}{4} tight)$$

2. **Acceleration ($a$)** is the derivative of velocity:

$$a(t) = \frac{dv}{dt} = -\left(\frac{\pi}{4}ight)^2 \sin\left(\frac{\pi}{4} tight) = -\frac{\pi^2}{16} \sin\left(\frac{\pi}{4} tight)$$

Both methods yield the exact same elegant expression for acceleration!

---

### Final Calculation

Now, we are ready to find the acceleration at the specific instant $t = 4/3	ext{ s}$.

Let's substitute $t = 4/3$ into our acceleration equation:

$$a\left(\frac{4}{3}ight) = -\frac{\pi^2}{16} \sin\left(\frac{\pi}{4} 	imes \frac{4}{3}ight)$$

Notice how beautifully the number $4$ cancels out in the numerator and denominator inside the sine function:

$$a\left(\frac{4}{3}ight) = -\frac{\pi^2}{16} \sin\left(\frac{\pi}{3}ight)$$

The angle $\frac{\pi}{3}$ radians is equivalent to $60^\circ$.

We know from standard trigonometry that:

$$\sin\left(\frac{\pi}{3}ight) = \frac{\sqrt{3}}{2}$$

Substituting this value back into our equation:

$$a\left(\frac{4}{3}ight) = -\frac{\pi^2}{16} 	imes \frac{\sqrt{3}}{2}$$

$$a\left(\frac{4}{3}ight) = -\frac{\sqrt{3}}{32} \pi^2 	ext{ cm s}^{-2}$$

This is our final answer, which perfectly matches **Option (d)**.

---

### Physical Significance of the Negative Sign

Let's take a moment to appreciate the physical meaning of the negative sign in our result.

At $t = 4/3	ext{ s}$, the displacement of the particle is:

$$x\left(\frac{4}{3}ight) = \sin\left(\frac{\pi}{3}ight) = \frac{\sqrt{3}}{2} \approx 0.866	ext{ cm}$$

Since the displacement is positive (the particle is to the right of the mean position), the restoring force must act in the opposite direction to pull it back towards the center.

Therefore, the acceleration must be negative, pointing back towards the origin.

This is the essence of simple harmonic motion—the system always fights to restore equilibrium!

Animated Solution for Physics - Oscillations: The $x - t$ graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at $t = 4/3 s$ is

Select Answer:

Visualized Solution

Analyzing the $x - t$ Graph

Finding Angular Frequency $ω$

Formulating the Displacement Equation

Relating Acceleration to Displacement

Setting up the Acceleration Equation

Substituting the Target Time $t = \frac{4}{3} s$

Evaluating the Sine Term

Calculating the Final Acceleration

Physical Significance & Insights

The Sigma Insight: Simple Harmonic Motion (SHM)

Introduction to Simple Harmonic Motion (SHM) Graphs

Analyzing the Setup

The Master Equation

Finding the Acceleration

Final Calculation

Physical Significance of the Negative Sign

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Animated Solution for Physics - Oscillations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The x−t graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at t=4/3 s is