Test: The Speed of a Travelling Wave & Principle of Superposition of Waves (September 28) - NEET MCQ

A traveling wave is a wave that moves through space or a medium and propagates in a particular direction. It can be represented by an equation of the form y = f(x - vt) or y = f(x + vt), where x is the spatial coordinate, t is time, v is the wave velocity, and f is a function of x and t.

Option A does not represent a traveling wave because it has an exponential term with x, which is not a characteristic of a traveling wave.

Option B also does not represent a traveling wave because it has an exponential term with x squared, which is not a characteristic of a traveling wave.

Option C represents a traveling wave because it has the form y = Asin(15x - 2t), where the wave is propagating in the positive x-direction with a speed of 2/15 m/s.

Therefore, the correct answer is Option C.

Explanation:

- The wave equation given is y = 0.021 sin (x + 30t), which is in the form of y = A sin(kx - ωt) where A is the amplitude, k is the wave number, and ω is the angular frequency.
- From the given wave equation, we can identify that ω = 30 rad/sec.
- The linear density (μ) of the string is given as 1.3 x 10^-3 kg/m.
- The tension (T) in the string can be found using the formula ω² = T/μ.
- Substituting the given values into the formula, we get T = ω² * μ.

Calculation:

- Substituting the given values, we get T = (30²) * (1.3 x 10^-3) kg m/s².
- Simplifying this, we get T = 0.12 N.

So, the tension in the string is 0.12 Newtons. Thus, the correct answer is A: 0.12 N.

Explanation:

To find the wavelength of a wave, we use the formula:

Wavelength = Speed / Frequency
 

• First, we need to plug in the given values into the formula. The speed (v) is given as 300 m/s and the frequency (f) is 60,000 Hz (or 60,000 s^-1).


• So, the formula becomes Wavelength = 300 / 60,000.


• Upon calculating this, we get the wavelength as 0.005 m.


• However, the options are given in centimeters. To convert meters to centimeters, we multiply by 100 (since 1 meter = 100 centimeters).


• So, 0.005 m * 100 = 0.5 cm.

Conclusion:

Therefore, the wavelength of the wave that is produced in air and travels at a speed of 300 m/s at the frequency of 60,000 Hz is 0.5 cm or Option A.

For more detailed explanations and practice problems, you can refer to the wave physics modules on EduRev.

**Understanding the Wave Equation**

The wave equation you provided, y = 15 x 10^-² sin (300t – 100x), is a standard equation for a wave propagating in the x direction.

It can be written in the form: y = A sin(ωt - kx), where:

- A is the amplitude of the wave,
- ω (omega) is the angular frequency,
- t is time,
- k is the wave number,
- x is the position.

**Finding the Wave Velocity**

The velocity (v) of a wave is related to its frequency (f) and wavelength (λ) by the equation v = fλ.

However, in our case, we don't have the frequency and the wavelength directly. We have the angular frequency (ω) and wave number (k) instead. But they are related to the frequency and wavelength by the equations ω = 2πf and k = 2π/λ.

So we can write the wave velocity as v = ω/k.

From the given wave equation, we can see that ω = 300 and k = 100.

So, the wave velocity v = ω/k = 300/100 = 3 m/s.

So, the correct answer is:
B: 3 m/s.

Understanding the Problem

• We are given the time it takes for the radar to detect the reflected waves from the plane.
• We are also given the speed of the waves.
• We need to find the distance of the plane from the radar.

Calculations

• The total time taken for the wave to travel to the airplane and back is given as 0.02 x 10^-3 seconds.
• Since the wave has to travel to the plane and back, the distance we want is half the total distance covered by the wave.
• The total distance covered by the wave is the speed of the wave multiplied by the total time taken, which is 3 x 10⁸ m/s x 0.02 x 10^-3 sec = 6000 meters.
• Therefore, the distance of the airplane from the radar is half of this, or 6000 meters / 2 = 3000 meters = 3 kilometers.

Answer
Therefore, the answer is C: 3.0 km. The plane is 3 kilometers away from the radar.

Please note: This solution is based on the assumption that the speed of the wave is constant and that it travels in a straight line to the plane and back. Actual conditions may vary.

This answer is provided by EduRev, a comprehensive educational platform that offers a wide range of learning resources for students of all levels.

Explanation

The answer is A: de-Broglie waves. Here's why:

de-Broglie waves

• de-Broglie waves, or matter waves, are associated with moving particles. The speed of de-Broglie waves is dependent on the momentum of the associated particle, and not on the speed of light. In fact, the de-Broglie wavelength is inversely proportional to the momentum of a particle. Thus, they do not necessarily travel at the speed of light.

X-rays

• X-rays are a type of electromagnetic radiation, like light waves. They have a much shorter wavelength than visible light, which enables them to penetrate materials that light cannot. But, like all electromagnetic waves, X-rays travel at the speed of light.

γ-rays (Gamma rays)

• Gamma rays are also a type of electromagnetic radiation, with even shorter wavelengths than X-rays. Due to their high energy, gamma rays can cause significant damage to living tissues and DNA. However, like all electromagnetic waves, gamma rays travel at the speed of light.

So, only de-Broglie waves among the options given do not travel with the speed of light.