When two shm y1= Asin wt & y2= a sin ( wt+π\3) in same direction.the resultant shm of the particle is

To find the resultant simple harmonic motion (SHM) when two SHM equations are given, we can use the principle of superposition. In your case, we have two SHM equations: \( y_1 = A \sin(\omega t) \) and \( y_2 = A \sin\left(\omega t + \frac{\pi}{3}\right) \). Let's break this down step by step to determine the resultant motion.

Understanding the Components

Both equations represent oscillations with the same amplitude \( A \) and angular frequency \( \omega \), but they are phase-shifted. The second equation has a phase shift of \( \frac{\pi}{3} \) radians, which means it starts oscillating a bit ahead of the first one.

Using the Superposition Principle

The principle of superposition states that when two or more waves overlap, the resultant displacement at any point is the sum of the individual displacements. Therefore, we can express the resultant \( y \) as:

• Resultant \( y = y_1 + y_2 \)

Substituting the equations, we get:

• Resultant \( y = A \sin(\omega t) + A \sin\left(\omega t + \frac{\pi}{3}\right) \)

Applying the Sine Addition Formula

To simplify this expression, we can use the sine addition formula. The formula states that:

• \( \sin(a + b) = \sin a \cos b + \cos a \sin b \)

Applying this to \( y_2 \):

• \( y_2 = A \left( \sin(\omega t) \cos\left(\frac{\pi}{3}\right) + \cos(\omega t) \sin\left(\frac{\pi}{3}\right) \right) \)

Since \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \) and \( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \), we can rewrite \( y_2 \) as:

• \( y_2 = A \left( \sin(\omega t) \cdot \frac{1}{2} + \cos(\omega t) \cdot \frac{\sqrt{3}}{2} \right) \)

Combining the Terms

Now, substituting \( y_2 \) back into the resultant equation:

• Resultant \( y = A \sin(\omega t) + A \left( \frac{1}{2} A \sin(\omega t) + \frac{\sqrt{3}}{2} A \cos(\omega t) \right) \)

This simplifies to:

• Resultant \( y = A \left( 1 + \frac{1}{2} \right) \sin(\omega t) + A \frac{\sqrt{3}}{2} \cos(\omega t) \)

Final Resultant Expression

Combining the coefficients gives us:

• Resultant \( y = \frac{3A}{2} \sin(\omega t) + \frac{\sqrt{3}}{2} A \cos(\omega t) \)

Now, we can express this in the form of a single SHM using the amplitude and phase shift. The resultant amplitude \( R \) can be calculated using:

• \( R = \sqrt{\left(\frac{3A}{2}\right)^2 + \left(\frac{\sqrt{3}}{2} A\right)^2} \)

Calculating this gives:

• \( R = A \sqrt{\frac{9}{4} + \frac{3}{4}} = A \sqrt{3} \)

Determining the Phase Angle

The phase angle \( \phi \) can be found using:

• \( \tan(\phi) = \frac{\text{Coefficient of } \cos(\omega t)}{\text{Coefficient of } \sin(\omega t)} = \frac{\frac{\sqrt{3}}{2} A}{\frac{3}{2} A} = \frac{\sqrt{3}}{3} \)

This gives us \( \phi = \frac{\pi}{6} \). Therefore, the resultant SHM can be expressed as:

• Resultant \( y = R \sin(\omega t + \phi) = A \sqrt{3} \sin\left(\omega t + \frac{\pi}{6}\right) \)

Summary of the Result

The resultant simple harmonic motion of the two given SHMs is:

• Resultant \( y = A \sqrt{3} \sin\left(\omega t + \frac{\pi}{6}\right) \)

This means that the two oscillations combine to form a new oscillation with a greater amplitude and a phase shift. This is a beautiful illustration of how waves can interact and combine in physics!