A particle osciallates with SHM according to the equation `x= (2.5 m ) cos [ ( 2pi t ) + (pi)/(4)]` . Its speed at `t = 1.5 ` s is

Options

A. `11.1 ms^(-1)`
B. `22.2 ms^(-1)`
C. `33.3ms^(-1)`
D. ` 44.4 ms^(-1)`

Answer

To find the speed of a particle oscillating in simple harmonic motion (SHM) given by the equation \( x = 2.5 \cos(2\pi t + \frac{\pi}{4}) \), we can follow these steps:

### Step 1: Differentiate the position function
The speed \( v \) of the particle is the derivative of the position \( x \) with respect to time \( t \):
\[ v = \frac{dx}{dt} \]
Given the equation:
\[ x = 2.5 \cos(2\pi t + \frac{\pi}{4}) \]
We differentiate this with respect to \( t \):
\[ v = -2.5 \cdot \sin(2\pi t + \frac{\pi}{4}) \cdot \frac{d}{dt}(2\pi t + \frac{\pi}{4}) \]
Since the derivative of \( 2\pi t + \frac{\pi}{4} \) with respect to \( t \) is \( 2\pi \), we have:
\[ v = -2.5 \cdot \sin(2\pi t + \frac{\pi}{4}) \cdot 2\pi \]
\[ v = -5\pi \sin(2\pi t + \frac{\pi}{4}) \]

### Step 2: Substitute \( t = 1.5 \) s into the speed equation
Now we substitute \( t = 1.5 \) s into the speed equation:
\[ v = -5\pi \sin(2\pi \cdot 1.5 + \frac{\pi}{4}) \]
Calculating the argument of the sine function:
\[ 2\pi \cdot 1.5 = 3\pi \]
Thus,
\[ v = -5\pi \sin(3\pi + \frac{\pi}{4}) \]

### Step 3: Simplify the sine function
Using the sine addition formula:
\[ \sin(a + b) = \sin a \cos b + \cos a \sin b \]
We can rewrite:
\[ \sin(3\pi + \frac{\pi}{4}) = \sin(3\pi) \cos(\frac{\pi}{4}) + \cos(3\pi) \sin(\frac{\pi}{4}) \]
Knowing that:
\[ \sin(3\pi) = 0 \quad \text{and} \quad \cos(3\pi) = -1 \]
\[ \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \quad \text{and} \quad \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \]
Thus,
\[ \sin(3\pi + \frac{\pi}{4}) = 0 \cdot \frac{1}{\sqrt{2}} + (-1) \cdot \frac{1}{\sqrt{2}} = -\frac{1}{\sqrt{2}} \]

### Step 4: Substitute back to find speed
Now substituting back:
\[ v = -5\pi \left(-\frac{1}{\sqrt{2}}\right) = \frac{5\pi}{\sqrt{2}} \]

### Step 5: Calculate the numerical value
Using \( \pi \approx 3.14 \):
\[ v \approx \frac{5 \times 3.14}{\sqrt{2}} \approx \frac{15.7}{1.414} \approx 11.1 \, \text{m/s} \]

Thus, the speed of the particle at \( t = 1.5 \) s is approximately \( 11.1 \, \text{m/s} \).

### Final Answer
The speed at \( t = 1.5 \) s is \( 11.1 \, \text{m/s} \).

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