A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm f

Q: Why did we ignore the negative sign in the acceleration formula?

The question specifically states that the **magnitude** of velocity is equal to the **magnitude** of acceleration. The negative sign in $a = -\omega^2 x$ simply indicates direction (towards the mean position), which we discard when taking the absolute value.

Q: Does the fact that the problem mentions 'SI units' change our calculation?

Surprisingly, no! If we convert the displacement from cm to meters, we divide both sides of our equation by 100. Since both velocity and acceleration have a linear dependence on length in their formulas, the conversion factor cancels out perfectly.

Q: Can we cancel $\omega$ from both sides of the equation? What if it's zero?

We can safely cancel $\omega$ because a particle executing Simple Harmonic Motion must have a non-zero angular frequency. If $\omega$ were zero, the particle would just be at rest, and there would be no oscillation!

Q: Why do we use $v = \omega \sqrt{A^2 - x^2}$ instead of $v = A\omega \cos(\omega t)$?

We use the position-dependent formula because the problem gives us the condition at a specific **position** ($x = 4$ cm), not at a specific **time**. Using the time-dependent formula would introduce an unknown variable $t$, making it impossible to solve directly.

Q: What does the angular frequency $\omega$ actually represent physically?

Angular frequency $\omega$ represents how fast the phase of the oscillation is changing. It's directly tied to the 'stiffness' of the restoring force relative to the mass. A higher $\omega$ means a stiffer system and a faster, more rapid oscillation.

Question

A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time (in seconds) is

Options:

(A) $\frac{4\pi}{3}$

(B) $\frac{8\pi}{3}$

(C) $\frac{7\pi}{3}$

(D) $\frac{3\pi}{8}$

Sigma Percentile · Accepted Answer

### Visualizing the Oscillator

Imagine a particle gracefully dancing back and forth along a straight line. This is the essence of **Simple Harmonic Motion (SHM)**.

The particle oscillates around a central point called the mean position ($x = 0$). It swings out to a maximum distance, which we call the amplitude ($A$). In our specific problem, we are given that this amplitude is $A = 5 	ext{ cm}$.

Our focus, however, is on a very specific snapshot in time: the exact moment when the particle is at a distance of $x = 4 	ext{ cm}$ from the mean position.

### The Kinematic Bridge

To solve this, we need to connect the particle's position to its velocity and acceleration. In SHM, we have powerful formulas that do exactly this without needing to know the time $t$.

The velocity $v$ of the particle at any position $x$ is given by the relation:
$v = \omega \sqrt{A^2 - x^2}$

Similarly, the acceleration $a$ is always directed towards the mean position and is proportional to the displacement:
$a = -\omega^2 x$

Here, $\omega$ is the **angular frequency**, a crucial parameter that dictates how rapidly the system oscillates.

### The Mathematical Duel

The problem presents a fascinating condition: at $x = 4 	ext{ cm}$, the magnitude of the velocity is exactly equal to the magnitude of the acceleration.

Let's set up our mathematical duel by equating their absolute values:
$|v| = |a|$

Substituting our kinematic formulas into this condition, we get:
$\omega \sqrt{A^2 - x^2} = |-\omega^2 x|$

Now, we plug in our known values, $A = 5$ and $x = 4$:
$\omega \sqrt{5^2 - 4^2} = \omega^2 (4)$

Let's simplify the square root. We know that $5^2 = 25$ and $4^2 = 16$. Their difference is $9$, and the square root of $9$ is simply $3$.

This beautifully reduces our complex equation to:
$3\omega = 4\omega^2$

Since the particle is actively oscillating, we know that $\omega$ cannot be zero. Therefore, we can safely divide both sides by $\omega$, yielding:
$\omega = \frac{3}{4} 	ext{ rad/s}$

### The Final Countdown

We have successfully found the angular frequency, but the question asks for the **periodic time** (or time period) $T$.

The bridge connecting angular frequency to the time period is one of the most fundamental relations in oscillatory physics:
$T = \frac{2\pi}{\omega}$

Let's substitute our newly found $\omega$ into this formula:
$T = \frac{2\pi}{3/4}$

When we divide by a fraction, we multiply by its reciprocal. The $4$ flips up to multiply with the $2\pi$:
$T = \frac{8\pi}{3} 	ext{ s}$

And there we have it! The time period of the oscillation is $\frac{8\pi}{3}$ seconds.

Animated Solution for Physics - Oscillations: A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time (in seconds) is

Select Answer:

Visualized Solution

$S H M S e t u p$

$V e l oc i t y an d A cce l er a t i o n$

$A ppl y in g t h e C o n d i t i o n$

$S im pl i f y in g t h e E q u a t i o n$

$S o l v in g f or ω$

$T im e P er i o d F or m u l a$

$F ina l C a l c u l a t i o n$

$T h e W a y F or w a r d$

The Sigma Insight: Simple Harmonic Motion

Visualizing the Oscillator

The Kinematic Bridge

The Mathematical Duel

The Final Countdown

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Animated Solution for Physics - Oscillations: A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time (in seconds) is

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SHM Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Velocity and Acceleration

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying the Condition

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying the Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for ω

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Time Period Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward