A particle executes S.H.M. given by the equation y = 0.45 sin 2t where y is in meter and t is in second. What is the speed of the particle when its displacement is 7.5 cm?

Options

A. 0.075`sqrt(3)ms^(-1)`
B. 7.5`sqrt(3)ms^(-1)`
C. 0.15 `sqrt(3)ms^(-1)`
D. 15`sqrt(3)ms^(-1)`

Answer

To solve the problem step by step, we will follow the process of analyzing the given equation of simple harmonic motion (S.H.M.), converting units, and applying the formulas for speed in S.H.M.

### Step 1: Understand the given equation
The equation of motion is given as:
\[ y = 0.45 \sin(2t) \]
where \( y \) is the displacement in meters and \( t \) is the time in seconds. The amplitude \( A \) of the motion is 0.45 m.

### Step 2: Convert the displacement into meters
The displacement we are interested in is 7.5 cm. We need to convert this into meters:
\[ 7.5 \text{ cm} = \frac{7.5}{100} \text{ m} = 0.075 \text{ m} \]

### Step 3: Find the sine value
We can set the displacement equal to the equation of motion to find \( \sin(2t) \):
\[ 0.075 = 0.45 \sin(2t) \]
Now, we can solve for \( \sin(2t) \):
\[ \sin(2t) = \frac{0.075}{0.45} = 0.1667 \]

### Step 4: Calculate \( \cos(2t) \)
Using the identity \( \cos^2(2t) + \sin^2(2t) = 1 \):
\[ \cos^2(2t) = 1 - \sin^2(2t) \]
\[ \cos^2(2t) = 1 - (0.1667)^2 \]
\[ \cos^2(2t) = 1 - 0.0278 = 0.9722 \]
Thus,
\[ \cos(2t) = \sqrt{0.9722} \approx 0.986 \]

### Step 5: Use the speed formula
The speed \( v \) in S.H.M. is given by:
\[ v = A \omega \cos(2t) \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency. From the equation, we see that:
- \( A = 0.45 \) m
- \( \omega = 2 \) rad/s

Now substituting the values:
\[ v = 0.45 \times 2 \times 0.986 \]
\[ v = 0.9 \times 0.986 \]
\[ v \approx 0.887 \text{ m/s} \]

### Step 6: Final result
The speed of the particle when its displacement is 7.5 cm is approximately:
\[ v \approx 0.887 \text{ m/s} \]