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Question #43859

Difficulty: 8/10 | 🟢 Free

The Question

A uniform heavy rope of mass $M$ and length $L$ hangs vertically from a rigid support. A transverse wave pulse is generated at its lower end and takes time $t_2$ to reach the top support. Now, a block of mass $m$ is attached to the lower end of the same rope. A new transverse wave pulse is generated at the lower end and takes time $t_1$ to reach the top support. If $t_1 = \frac{1}{2} t_2$ , what is the ratio $\frac{m}{M}$ ? (Assume the rope does not stretch and the transverse pulses are small).

Options

A.
$\frac{9}{16}$
B.
$\frac{3}{2}$
C.
3
D.
1

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

String Tension and Wave Speed|Effective Spring Constant|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to model a continuous medium with variable local properties, specifically the non-linear time-of-flight of a wave pulse under gravity.

A common trap is treating the string as having a constant effective tension (either the average tension or the maximum tension at the support) rather than integrating the infinitesimal time elements $dt = dy/v(y)$ .

For instance, using the average tension $T_{avg} = mg + Mg/2$ leads to a constant assumed velocity and results in the incorrect ratio $3/2$ .

Another trap is misapplying the binomial approximation for large mass ratios, leading to $1$ .

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Waves - Class 11 Part 2
Concept	String Tension and Wave Speed
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free