A particle performs simple harmonic motion with a period of 2 s. The time taken by the particle to c

Q: Why did we use the sine function and not the cosine function?

The choice of trigonometric function depends entirely on the initial conditions. Since the particle starts from the mean position ($x=0$ at $t=0$), the sine function is appropriate because $\sin(0) = 0$. If it had started from the extreme position ($x=A$ at $t=0$), we would use the cosine function because $\cos(0) = 1$.

Q: What if the particle started from the extreme position? How much time would it take to reach $A/2$?

If it started from the extreme position, the equation would be $x = A \cos(\omega t)$. Substituting $x = A/2$, we get $\cos(\omega t) = 1/2$, which means $\omega t = \pi/3$. This gives $t = T/6$. In this specific problem where $T=2$ s, it would take $1/3$ seconds.

Q: Why did we take the angle as $\pi/6$ and not $5\pi/6$?

The equation $\sin(\theta) = 1/2$ has multiple solutions: $\pi/6$, $5\pi/6$, etc. However, we are looking for the *first* time the particle reaches $A/2$ after starting from the mean position. Therefore, we must choose the smallest positive angle, which is $\pi/6$.

Q: Does the time taken depend on the amplitude $A$?

No, it does not. As you can see in the derivation, the amplitude $A$ cancels out from both sides of the equation. The time taken to reach a specific *fraction* of the amplitude depends only on the time period $T$.

Q: Can we solve this using the phase circle (phasor) method?

Absolutely! In the phasor diagram, starting from the mean position means starting at an angle of $0^\circ$. Reaching $x = A/2$ corresponds to a vertical projection of $A/2$, which means the phasor has rotated by an angle $\theta$ where $\sin(\theta) = (A/2)/A = 1/2$. Thus, $\theta = 30^\circ$ or $\pi/6$ radians. Since $2\pi$ radians take $T$ seconds, $\pi/6$ radians will take $T/12$ seconds. With $T=2$ s, $t = 2/12 = 1/6$ s.

Question

A particle performs simple harmonic motion with a period of 2 s. The time taken by the particle to cover a displacement equal to half of its amplitude from the mean position is $1/a$ s. The value of $a$ to the nearest integer is ......... .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a particle executing Simple Harmonic Motion (SHM). The problem states that the particle starts from its mean position and travels to a displacement equal to half of its amplitude. We are given the time period of the oscillation, $T = 2 	ext{ s}$, and we need to find the exact time $t$ it takes to reach $x = \frac{A}{2}$.

Whenever a particle begins its journey from the mean position ($x = 0$) and moves towards the positive extreme, its displacement as a function of time is perfectly modeled by the sine function:

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

Here, $A$ is the maximum amplitude, and $\omega$ is the angular frequency. We also know the fundamental relationship between angular frequency and time period:

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

### The Master Equation

Let's substitute the known values into our displacement equation. We know the target displacement is $x = \frac{A}{2}$ and the time period is $T = 2 	ext{ s}$. Plugging these in, we get:

$$\frac{A}{2} = A \sin\left(\frac{2\pi}{2} tight)$$

Notice how beautifully the equation simplifies. The amplitude $A$ cancels out from both sides, which tells us a profound physical truth: **the time taken to reach a specific fraction of the amplitude is completely independent of the amplitude itself!**

Furthermore, the $2$ in the numerator and denominator of the angle also cancel out, leaving us with a clean, purely mathematical trigonometric equation:

$$\frac{1}{2} = \sin(\pi t)$$

### Final Calculation

Now, we must ask ourselves: at what angle does the sine function equal $\frac{1}{2}$? Since the particle is reaching this point for the very first time, we look for the smallest positive angle in the first quadrant. From our standard trigonometric values, we know that $\sin\left(\frac{\pi}{6}ight) = \frac{1}{2}$.

Equating the angles, we get:

$$\pi t = \frac{\pi}{6}$$

Solving for $t$, the $\pi$ cancels out, yielding:

$$t = \frac{1}{6} 	ext{ s}$$

The problem states that this time is equal to $\frac{1}{a} 	ext{ s}$. By direct comparison:

$$\frac{1}{a} = \frac{1}{6}$$

Therefore, the value of $a$ is exactly **$6$**.

Always remember to check the starting position in SHM problems. If the particle had started from the extreme position, we would have used the cosine function, $x(t) = A \cos(\omega t)$, and the time to reach $A/2$ would have been $T/6$ instead of $T/12$.

Animated Solution for Physics - Oscillations: A particle performs simple harmonic motion with a period of 2 s. The time taken by the particle to cover a displacement equal to half of its amplitude from the mean position is $1/ a$ s. The value of $a$ to the nearest integer is ......... .

Enter Numerical Value:

Visualized Solution

$x = 0 \to x = A /2$

$Equation of SHM$

$Substitution$

$Simplification$

$Solving for t$

$Finding a$

$The Way Forward$

The Sigma Insight: Simple Harmonic Motion (SHM)

Analyzing the Setup

The Master Equation

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0→x=A/2

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

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

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

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

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

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