Waves And Oscillations

Step 1: Form the resultant wave.
The superposition of the two waves gives: At :

Step 2: Determine when velocity is zero.
Velocity of the particle: The velocity becomes zero when .

Step 3: Use given time intervals.
Velocity is zero at . The time difference between consecutive zeros:

Step 4: Calculate frequency.

Step 5: Conclusion.
Hence, the minimum frequency of the waves is .

Step 1: Use SHM velocity equation.
At ; At .

Step 2: Take ratio.

Step 3: Find .

Step 4: Conclusion.
Hence, and .

Step 1: Represent the waves as phasors.
The resultant wave amplitude is obtained from vector addition of phasors. Let , , .

Step 2: Calculate net amplitude.

Step 3: Determine phase.

Step 4: Conclusion.
Thus, and .

To solve the problem, let's analyze the given simple harmonic motions along the x and y axes:

For the x-axis motion:

This can be simplified using the trigonometric identity :

For the y-axis motion:

We will use the parametric equations for the x and y motions to find the resultant equation of motion. Combining both equations:

Squaring both equations and adding:

Recall the double angle identity: .

Substituting this into x²:

Thus, we have:

Substitute into the equation for x²:

Recognizing that:

Leads to:

This corresponds to the resultant motion , which matches option 4. This is the path described by the particle, which typically represents an ellipse.

To solve this problem, we need to understand the concepts of simple harmonic motion (SHM) and how displacement ( ), velocity ( ), and acceleration ( ) relate to each other.

1. Understanding Simple Harmonic Motion Equations:
   • The displacement in SHM is given by:
   • The velocity is the derivative of the displacement:
   • The acceleration is the derivative of the velocity:
2. Relating Acceleration and Displacement:
   • From the equations, acceleration can also be expressed as:
   • This implies , a constant.
   • Given the period , we have .
   • Thus, , which does not change with time.
3. Relating Velocity and Acceleration:
   • Using earlier expressions: and .
   • Substitute into the acceleration equation.
   • It can be shown that the equation connecting and forms an ellipse equation upon squaring and adding.
4. Conclusion:
   • The correct answers are:
     1. does not change with time.
     2. and are related by an equation of an ellipse.

Given:
Time period when elevator is stationary:
Elevator accelerates upward with
Effective gravity:

Time period of a simple pendulum: Thus, Required ratio: Calculate: Final Answer:

To determine which statements are correct regarding the Fourier series expansion of the periodic function in the interval , let's go through the process of Fourier series expansion step-by-step.

Step 1: Fourier Series Expansion

A Fourier series for a function in the interval to is given by:

Where:

Step 2: Calculation of Coefficients

First, let's determine the term:

Thus, the first term is:

Next, let's calculate :

This integral evaluates to zero because the integrand is an odd function over a symmetric interval about the origin.

Now, let's determine :

The integral for generally involves integration by parts and yields:

Step 3: Forming the Series

The Fourier series expansion becomes:

Conclusion

Given these calculations:

• Coefficients of all the sine terms are zero: Correct, as explained .
• The first term in the series is : Correct, as shown above.
• The second term in the series is -4cosx: Correct as was calculated to be -4.
• Coefficients of all the cosine terms are zero: Incorrect, since the cosine terms do exist and are significant in the expansion.

Velocity Component Along the Line of Sight

The velocity component v_PS(t) along the line joining the source and the observer is given by:

v_PS(t) = v(t) cosϕ

where ϕ is the angle between the velocity vector and the line joining S and P.

Distance Between the Source and Observer

From the geometry of the problem, the distance between the source and the observer is:

r(t) = √(x(t)² + L²) = √(a² sin²(ωt) + L²)

The angle ϕ satisfies:

cosϕ = L / r(t)

Velocity Component Substitution

Substituting v(t) = v₀ cos(ωt) and cosϕ = L / r(t), we have:

v_PS(t) = v₀ cos(ωt) × (L / √(a² sin²(ωt) + L²))

Using Trigonometric Identities

The component along the line of sight becomes:

v_PS(t) = (1/2) × (a v₀ / √(a² sin²(ωt) + L²)) sin(2ωt)

Thus, statement (A) is correct.

Doppler Effect and Observed Frequency

The observed frequency is given by the Doppler effect, which depends on the relative velocity component v_PS(t).

• At x = 0 (midpoint of oscillation) and x = ±a (extremes of oscillation), the relative velocity component becomes zero, and the observed frequency equals the original frequency f.

Hence, statement (B) is correct.

Analyzing Other Statements

• At x = ±a/2, the velocity component v_PS(t) is nonzero, so the observed frequency differs from f.
• Statement (D) is incorrect because v_PS(t) depends on the varying distance r(t), not just √(a² + L²).

Conclusion

The correct statements are (A) and (B).

When the two waves and are superposed, the resultant wave can be expressed as a combination of a carrier wave and a modulation wave. The carrier frequency is determined by the average of the angular frequencies, while the modulation frequency is determined by their difference.

Given:

Resultant Wave:

The resultant wave is given by the superposition of and :

• The carrier frequency is determined by the term .
• The modulation frequency is determined by the term .

Substitute :

For , the carrier frequency becomes:

Conclusion:

Thus, the correct statement is that the carrier frequency of the resultant wave is .

The given equation represents a quadratic function of x, which suggests a parabolic trajectory. This is characteristic of projectile motion under the influence of gravity, where the path traced by an object is parabolic if the acceleration due to gravity is constant and the object is projected with some initial velocity.
Thus, the correct answer is (A) Projectile motion in a uniform gravitational field.

For a particle moving through a potential with a given energy, the probability density represents the likelihood of finding the particle at a given position. The potential that produces the required probability distribution in the figure should have the corresponding characteristics based on the energy and the nature of the particle’s behavior.

Option (C) matches the required potential shape.