← 2019 Paper 1

UPSC 2019 Maths Optional Paper 1 Q7c — Step-by-Step Solution

20 marks · Section B

Simple harmonic motion (free, damped, forced) · Dynamics & Statics · asked 8× in 13 yrs · Read the full method →

Question

A particle moving along the $y$-axis has an acceleration $Fy$ towards the origin, where $F$ is a positive and even function of $y$. The periodic time, when the particle vibrates between $y=-a$ and $y=a$, is $T$. Show that

where $F_1$ and $F_2$ are the greatest and the least values of $F$ within the range $[-a,a]$. Further, show that when a simple pendulum of length $l$ oscillates through $30°$ on either side of the vertical line, $T$ lies between $2\pi\sqrt{l/g}$ and $2\pi\sqrt{l/g}\sqrt{\pi/3}$.

Technique

Energy first integral $\dot y^2=2\int_y^a Fs\,ds$; bound $F$ by its extremes $F_1,F_2$ inside the period integral; the bounding integrals reduce to $\int_0^a dy/\sqrt{a^2-y^2}=\pi/2$. For the pendulum, recast $g\sin\theta$ as $F\cdot y$ with $F=(g/l)(\sin\theta/\theta)$, then read off $F_1,F_2$ from the monotonicity of $\sin\theta/\theta$.

Solution

Step 1 — Equation of motion and energy integral

The acceleration is $\ddot y=-F(y)\,y$ (toward the origin). Multiply by $\dot y$ and integrate:

Integrating from the extreme position $y=a$ (where $\dot y=0$) to a general $y$:

Step 2 — Period as an integral

By symmetry (quarter period from $y=0$ to $y=a$), with $\dot y=-\sqrt{\dot y^2}$ on the inward leg,

Step 3 — Sandwich the inner integral

Let $F_1=\max_{[-a,a]}F$ and $F_2=\min_{[-a,a]}F$ (both positive). For every $s$,

and $\displaystyle\int_y^a s\,ds=\tfrac12(a^2-y^2)$. Hence

Taking $\sqrt{2(\cdot)}$ and inverting (larger denominator $\Rightarrow$ smaller integrand):

Step 4 — Integrate the bounds

Using $\displaystyle\int_0^a\frac{dy}{\sqrt{a^2-y^2}}=\Big[\sin^{-1}\tfrac ya\Big]_0^a=\frac\pi2$, multiply by $4$:

(The inequalities are strict because $F$ is non-constant on $[-a,a]$, so the bounding is strict on a set of positive measure.)

Step 5 — The simple pendulum oscillating through $30^\circ$

For a pendulum of length $l$, with $\theta$ the angular displacement, the tangential equation is $l\ddot\theta=-g\sin\theta$. Write the arc displacement $y=l\theta$ (so the particle moves on the “$y$-axis” of arc length). Then

Thus $F(y)=\dfrac gl\cdot\dfrac{\sin\theta}{\theta}$ with $y=l\theta$, and $F$ is positive and even in $y$ (since $\sin\theta/\theta$ is even). The amplitude is $\theta=\pi/6$ ($30^\circ$), so $y$ ranges over $[-a,a]$ with $a=l\pi/6$.

On $[0,\pi/6]$, $\dfrac{\sin\theta}{\theta}$ is decreasing, so:

• Greatest $F$: at $\theta\to0$, $\dfrac{\sin\theta}{\theta}\to1$, giving $F_1=\dfrac gl$.
• Least $F$: at $\theta=\pi/6$, $\dfrac{\sin(\pi/6)}{\pi/6}=\dfrac{1/2}{\pi/6}=\dfrac{3}{\pi}$, giving $F_2=\dfrac gl\cdot\dfrac3\pi$.

Apply Step 4:

Therefore

Answer