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Question #43499

Difficulty: 8/10 | 🟢 Free

The Question

A uniform heavy spring of natural length $L$ , mass $M$ , and stiffness constant $k$ is suspended vertically from a rigid support. A block of mass $m$ is attached to its lower end. The system is allowed to reach equilibrium. The block is then pulled down slightly and released, performing vertical simple harmonic oscillations. Assuming that the spring stretches uniformly along its length at all times, what is the time period $T$ of the oscillations?

Options

A.
$2\pi \sqrt{\frac{3m + M}{3k}}$
B.
$2\pi \sqrt{\frac{m + M}{k}}$
C.
$2\pi \sqrt{\frac{2m + M}{2k}}$
D.
$2\pi \sqrt{\frac{m}{k}}$

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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

Tension in Massless vs Massive Strings|SHM Definition via Differential Equation|Effective Spring Constant

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to transition from discrete to continuous mass systems in SHM using the energy method.

The primary trap is incorrectly assuming the entire mass of the heavy spring moves at the same velocity as the attached block, leading to an effective mass of $m + M$ .

Another common mistake is assuming the effective mass is $m + M/2$ by treating the spring's center of mass as the representative velocity point.

By forcing the derivation of the kinetic energy integral $K = \int \frac{1}{2} (dm) v(y)^2$ , the student must recognize the linear velocity profile $v(y) = (y/L)v$ , which yields the correct effective mass addition of $M/3$ .

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Laws of Motion - Class 11 Part 1
Concept	Tension in Massless vs Massive Strings
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free