The amplitude of given simple harmonic motion is `y=(3sinomegat+4cosomegat)m`

Options

A. 1 m
B. 3 m
C. 4 m
D. 5 m

Answer

To find the net amplitude of the given simple harmonic motion represented by the equation \( y = 3 \sin(\omega t) + 4 \cos(\omega t) \), we can follow these steps:

### Step 1: Identify the components of the motion
The equation consists of two components:
- \( y_1 = 3 \sin(\omega t) \)
- \( y_2 = 4 \cos(\omega t) \)

### Step 2: Convert one component to sine form
We can express \( y_2 \) in terms of sine. We know that:
\[ \cos(\theta) = \sin\left(\theta + 90^\circ\right) \]
Thus, we can rewrite \( y_2 \):
\[ y_2 = 4 \cos(\omega t) = 4 \sin\left(\omega t + 90^\circ\right) \]

### Step 3: Analyze the phase difference
Now we have:
- \( y_1 = 3 \sin(\omega t) \)
- \( y_2 = 4 \sin\left(\omega t + 90^\circ\right) \)

The phase difference between \( y_1 \) and \( y_2 \) is \( 90^\circ \).

### Step 4: Use the Pythagorean theorem to find the resultant amplitude
Since the two components are perpendicular to each other (due to the \( 90^\circ \) phase difference), we can find the net amplitude \( A \) using the Pythagorean theorem:
\[ A = \sqrt{(3)^2 + (4)^2} \]
Calculating this gives:
\[ A = \sqrt{9 + 16} = \sqrt{25} = 5 \, \text{m} \]

### Step 5: Conclusion
Thus, the net amplitude of the given simple harmonic motion is \( 5 \, \text{m} \).

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