Equation of SHM is `x = 10 sin 10 pi t`. Find the distance between the two points where speed is `50pi cm//sec`. x is in cm and t is in seconds

Options

A. zero
B. 20cm
C. 17.32 cm
D. 8.66 cm

Answer

To solve the problem, we need to find the distance between two points where the speed of the simple harmonic motion (SHM) is \(50\pi \, \text{cm/s}\). The equation of SHM is given as:

\[ x = 10 \sin(10 \pi t) \]

### Step 1: Identify the parameters
From the equation \(x = 10 \sin(10 \pi t)\):
- Amplitude \(A = 10 \, \text{cm}\)
- Angular frequency \(\omega = 10\pi \, \text{rad/s}\)

### Step 2: Write the expression for velocity
The velocity \(v\) in SHM can be expressed as:
\[ v = \omega \sqrt{A^2 - x^2} \]

### Step 3: Substitute known values
We know that \(v = 50\pi \, \text{cm/s}\). Substituting the values of \(\omega\) and \(A\) into the velocity equation:
\[ 50\pi = 10\pi \sqrt{100 - x^2} \]

### Step 4: Simplify the equation
Dividing both sides by \(10\pi\):
\[ 5 = \sqrt{100 - x^2} \]

### Step 5: Square both sides
Squaring both sides to eliminate the square root gives:
\[ 25 = 100 - x^2 \]

### Step 6: Solve for \(x^2\)
Rearranging the equation:
\[ x^2 = 100 - 25 = 75 \]

### Step 7: Find the values of \(x\)
Taking the square root of both sides:
\[ x = \pm \sqrt{75} = \pm 5\sqrt{3} \]

### Step 8: Calculate the distance between the two points
The two points where the speed is \(50\pi \, \text{cm/s}\) are \(x_1 = 5\sqrt{3}\) and \(x_2 = -5\sqrt{3}\). The distance \(d\) between these two points is:
\[ d = x_1 - x_2 = 5\sqrt{3} - (-5\sqrt{3}) = 5\sqrt{3} + 5\sqrt{3} = 10\sqrt{3} \]

### Step 9: Calculate the numerical value
Using the approximate value of \(\sqrt{3} \approx 1.732\):
\[ d = 10 \times 1.732 = 17.32 \, \text{cm} \]

### Conclusion
The distance between the two points where the speed is \(50\pi \, \text{cm/s}\) is \(17.32 \, \text{cm}\).