A transverse wave on a string has an amplitude of 0.2 m and a frequency of 175 Hz. Consider a particle of the string at x=0. it begins with a displacement y=0, at t=0, according to equation `y=0.2 sin (kx+-omegat)`. How much time passed between the first two instant when this particle has a displacement of y=0.1 m?

Options

A. 1.9 ms
B. 3.9 ms
C. 2.4 ms
D. 0.5 ms

Explanation

To solve the problem, we need to find the time interval between the first two instances when a particle of the string has a displacement of \( y = 0.1 \, \text{m} \). The wave equation is given as:

\[ y = 0.2 \sin(kx - \omega t) \]

### Step 1: Set up the equation for displacement
Since we are considering the particle at \( x = 0 \), the equation simplifies to:

\[ y = 0.2 \sin(-\omega t) \]

We need to find when \( y = 0.1 \, \text{m} \):

\[ 0.1 = 0.2 \sin(-\omega t) \]

### Step 2: Solve for sine
Dividing both sides by 0.2 gives:

\[ \sin(-\omega t) = \frac{0.1}{0.2} = \frac{1}{2} \]

### Step 3: Find the angles for sine
The sine function equals \( \frac{1}{2} \) at:

\[ -\omega t = \frac{\pi}{6} \quad \text{and} \quad -\omega t = \frac{5\pi}{6} \]

### Step 4: Solve for time
From the first equation:

\[ \omega t_1 = -\frac{\pi}{6} \implies t_1 = -\frac{\pi}{6\omega} \]

From the second equation:

\[ \omega t_2 = -\frac{5\pi}{6} \implies t_2 = -\frac{5\pi}{6\omega} \]

### Step 5: Calculate the time difference
The time difference \( \Delta t \) between the two instances is:

\[ \Delta t = t_2 - t_1 = \left(-\frac{5\pi}{6\omega}\right) - \left(-\frac{\pi}{6\omega}\right) \]

This simplifies to:

\[ \Delta t = -\frac{5\pi}{6\omega} + \frac{\pi}{6\omega} = -\frac{4\pi}{6\omega} = -\frac{2\pi}{3\omega} \]

### Step 6: Substitute for angular frequency
The angular frequency \( \omega \) is related to the frequency \( f \) by the formula:

\[ \omega = 2\pi f \]

Given that the frequency \( f = 175 \, \text{Hz} \):

\[ \omega = 2\pi \times 175 \]

Substituting \( \omega \) into the time difference equation:

\[ \Delta t = -\frac{2\pi}{3 \cdot (2\pi \times 175)} = -\frac{2}{3 \cdot 350} = -\frac{2}{1050} = -\frac{1}{525} \, \text{s} \]

### Step 7: Convert to milliseconds
To convert seconds to milliseconds:

\[ \Delta t = -\frac{1}{525} \times 1000 \approx -1.9 \, \text{ms} \]

Since we are interested in the time interval, we take the absolute value:

\[ \Delta t \approx 1.9 \, \text{ms} \]

### Final Answer
The time passed between the first two instances when the particle has a displacement of \( y = 0.1 \, \text{m} \) is approximately **1.9 milliseconds**.

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