Before You Begin

Question #43821

Difficulty: 8/10 | 🟢 Free

The Question

A mechanical oscillator consists of a block of mass $m$ (in kg) attached to a spring of force constant $k$ (in N/m). The block is rigidly connected to a rectangular conducting loop of width $L$ (in m) and total resistance $R$ (in $\Omega$ ). The loop moves in a uniform magnetic field $B$ (in T) perpendicular to its plane, such that only one side of length $L$ remains in the field. A radioactive sample with initial number of nuclei $N_0$ and decay constant $\lambda$ (in s $^{-1}$ ) is placed on the block. An external mechanism exerts a time-varying driving force on the block proportional to the instantaneous activity of the sample: $F_{ext}(t) = \alpha A(t)$ , where $A(t) = -\frac{dN}{dt}$ is the activity and $\alpha$ (in N $\cdot$ s) is a positive coupling constant. Assuming the system's displacement from equilibrium can be modeled with a particular solution of the form $x_p(t) = X_0 e^{-\lambda t}$ , what is the amplitude coefficient $X_0$ (in m)?

Options

A.
$\frac{\alpha \lambda N_0}{k + m\lambda^2 - \frac{B^2 L^2 \lambda}{R}}$
B.
$\frac{\alpha \lambda N_0}{k + m\lambda^2 + \frac{B^2 L^2 \lambda}{R}}$
C.
$\frac{\alpha \lambda N_0}{\sqrt{(k - m\lambda^2)^2 + \left(\frac{B^2 L^2 \lambda}{R}\right)^2}}$
D.
$\frac{\alpha \lambda N_0}{k + m\lambda^2}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Activity Units and Transitions|SHM Definition via Differential Equation|Lenz’s Law Energy Conservation

Why This Question Is Tricky

Common Mistakes

This question tests the misapplication of driven oscillator formulas.

Students often memorize the steady-state amplitude for a sinusoidally driven oscillator (which involves a quadrature sum in the denominator) or assume that damping always strictly 'adds' to the impedance of the system.

Here, the driving force is an exponential decay.

Because the particular solution $x_p(t)$ monotonically decays, its velocity is strictly negative while displacement is positive.

Consequently, the Lenz's Law damping force acts in the same direction as the driving force, effectively reducing the apparent stiffness of the system.

This leads to a negative sign for the damping term in the characteristic denominator, a counter-intuitive result for students who rely on rote memorization of AC circuits or harmonic SHM.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Nuclei - Class 12
Concept	Activity Units and Transitions
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free