Question

Consider a rope fixed at both ends under tension so that it is horizontal (i.e. assume the rope is along x-axis, with gravity acting along z-axis). Now the right end is continually oscillated at high frequency n (say n=100 Hz) horizontally and in a direction along the rope; amplitude of oscillation is negligible. The oscillation travells along the rope and is reflected at the left end. 
Let the total length of rope be l, total mass be m and the acceleration due to gravity be g. 
After initial phase (say a mintue or so), the rope has __(BLANK-1)__ wave, which is __(BLANK-2)__ in nature. It results from superposition of left travelling and right travelling __(BLANK-3)__ waves. This resulting wave has a frequency __ (BLANK-4)_ that of oscillation frequency nu. Simple dimensional analysis indicates that the frequency of can be of the form: ___(BLANK-5)__ .

• A. BLANK-1: travelling, oscillating, stationary, regular
• B. BLANK-2: transverse, longitudinal, regular, irregular
• C. BLANK-3: transverse, longitudinal, regular, irregular
• D. BLANK-4: equal to, half, double, independent from

Hint:

In a tensioned rope, waves propagate with velocity determined by the tension and mass per unit length. For a given wave, the frequency remains unchanged by reflection.

Answer 1

The Correct Option is A

Let's analyze each part based on the physics described:

Part 1: The Rope Wave

1. Wave Type Identification: The oscillation is applied horizontally *along* the direction of the rope (x-axis). The wave travels along the rope. When the particle oscillation direction is parallel to the wave propagation direction, the wave is defined as longitudinal.
2. Superposition and Reflection: The longitudinal wave travels to the left end, reflects (at a fixed end, a longitudinal wave typically reflects with a phase shift, but the key point is reflection occurs), and travels back to the right. The incident wave (traveling left) and the reflected wave (traveling right) have the same frequency and are confined between two fixed ends.
3. Resultant Wave: The superposition (interference) of two identical waves traveling in opposite directions in a confined medium results in a standing wave, also known as a stationary wave. This wave pattern does not propagate; it has fixed points of zero amplitude (nodes) and maximum amplitude (antinodes).

Part 2: The Simple Pendulum (Inferred from options for D and E)

1. Pendulum Period: The formula for the period (\(T\)) of a simple pendulum (for small oscillations) is \( T = 2\pi \sqrt{\frac{l}{g}} \), where \(l\) is the length and \(g\) is the acceleration due to gravity. Notice that the mass (\(m\)) of the pendulum bob does not appear in this formula.
2. Period vs. Mass: Therefore, the period of a simple pendulum is independent from its mass.
3. Pendulum Frequency: Frequency (\(n\) or \(f\)) is the reciprocal of the period: \( f = \frac{1}{T} = \frac{1}{2\pi} \sqrt{\frac{g}{l}} \). The frequency depends on \(g\) and \(l\). Specifically, the frequency squared (\(f^2\)) is directly proportional to the ratio \(g/l\). While the frequency itself is proportional to \(\sqrt{g/l}\), among the given options, \(g/l\) is the quantity that fundamentally determines the frequency (along with constants).

Filling in the Blanks:

• (A) BLANK-1: The resultant wave pattern is a standing wave. Correct choice: stationary
• (B) BLANK-2: The wave involves oscillations parallel to propagation. Correct choice: longitudinal
• (C) BLANK-3: Classifying the relationship between oscillation and propagation direction. Correct choice: longitudinal
• (D) BLANK-4: The relationship between a pendulum's period and its mass. Correct choice: independent from
• (E) BLANK-5: The quantity related to pendulum frequency dependence on \(g\) and \(l\). Correct choice: \(g/l\)

Answer 2

After the initial phase, the rope exhibits travelling transverse waves, which are a result of the superposition of the left and right traveling waves. The frequency of the resulting wave is equal to the frequency of the oscillating source. The dimensional analysis of wave velocity in such a system leads to the result \( \sqrt{\frac{g}{l}} \).

Answer 3

Let us analyze each blank based on the physics of wave motion in a rope: 

(A) The wave that travels along the rope due to the oscillation is a stationary wave. → BLANK-1: stationary

(B) Since the oscillation is along the direction of the rope (i.e. along the x-axis), the type of wave is longitudinal. → BLANK-2: longitudinal 

(C) The reflected wave retains its nature, so it is also longitudinal. → BLANK-3: longitudinal 

(D) Since the wave reflects back without change in medium, its frequency remains equal to the original. → BLANK-4: equal to 

(E) The natural frequency of a simple pendulum (or the characteristic frequency for small oscillations in systems under gravity) is given by: \[ f = \sqrt{\frac{g}{l}} \] → BLANK-5: \(\sqrt{\frac{g}{l}}\)