Simple Harmonic Motion

$ \begin{aligned} &\text { For the given spring mass system, }\\ &k_{\mathrm{eq}}=k+2 k+k=4 k \end{aligned} $

$ \text { ∴ Frequency, } \eta=\frac{1}{2 \pi} \sqrt{\frac{k_{\text {eq }}}{m}}=\frac{1}{2 \pi} \sqrt{\frac{4 k}{m}} $

To determine the time periods of vertical oscillations in different configurations of springs, we examine each setup.

Configuration (a):

The time period, $ T_a $, is given by the formula:

$ T_{a} = 2\pi \sqrt{\frac{m}{k}} $

Configuration (b):

In this setup, two identical springs are arranged in series, effectively halving the spring constant. Thus, the time period, $ T_b $, is:

$ T_{b} = 2\pi \sqrt{\frac{m}{\frac{k}{2}}} = 2\pi \sqrt{\frac{2m}{k}} $

Configuration (c):

In this setup, two identical springs are arranged in parallel, effectively doubling the spring constant. Hence, the time period, $ T_c $, is:

$ T_{c} = 2\pi \sqrt{\frac{m}{2k}} $

Now, calculating the ratios of these time periods, we get:

$ T_{a} : T_{b} : T_{c} = 1 : \sqrt{2} : \frac{1}{\sqrt{2}} $

This ratio expresses the relative oscillation times for each configuration of the springs.

The motion of the particle performing S.H.M. represented by

$\begin{aligned} x & =A \sin \omega t=1 \sin \frac{2 \pi}{T} \cdot t=\sin \frac{2 \pi}{8} t \\ \therefore \quad x & =\sin \left(\frac{\pi t}{4}\right) \end{aligned}$

$\therefore$ Velocity, $v=\frac{d x}{d t}$

$=\frac{d}{d t} \sin \left(\frac{\pi}{4}\right) t=\frac{\pi}{4} \cos \left(\frac{\pi}{4}\right) t$

$\therefore$ Acceleration of the particle.

$\begin{aligned} a & =\frac{d v}{d t}=\frac{d}{d t}\left[\frac{\pi}{4} \cos \left(\frac{\pi}{4}\right) t\right] \\ & =-\frac{\pi}{4} \cdot \frac{\pi}{4} \cdot \sin \left(\frac{\pi}{4} t\right) \Rightarrow a=-\frac{\pi^2}{16} \sin \left(\frac{\pi}{4} \cdot t\right) \end{aligned}$

At $t=2 \mathrm{~s}$

$a=-\frac{\pi^2}{16} \cdot \sin \left(\frac{\pi}{4} \cdot 2\right)=-\frac{\pi^2}{16} \mathrm{~ms}^{-2}$

Let k be the force constant of spring.

Time period, $T = 2\pi \sqrt {{M \over k}} $ ...... (i)

When block of mass m is placed in the tray, the time period of oscillations becomes

$T' = 2\pi \sqrt {{{M + m} \over k}} $ ...... (ii)

On dividing Eq. (ii) by Eq. (i), we get

$ \Rightarrow {{T'} \over T} = {{\sqrt {M + m} } \over {\sqrt M }}$

$ \Rightarrow {3 \over {1.5}} = \sqrt {{{12 + m} \over {12}}} $ [given, M = 12 kg, T = 1.5 s, T' = 3 s]

$ \Rightarrow \sqrt {{{12 + m} \over {12}}} = 2$

$\Rightarrow$ 12 + m = 48

$\Rightarrow$ m = 36 kg

Let displacement, x = A$\omega$ sin $\omega$t

Velocity, $v = {{dx} \over {dt}} = A\omega \cos \omega t$

Kinetic energy of particle in SHM,

$K = {1 \over 2}m{v^2} \Rightarrow K = {1 \over 2}m{\omega ^2}{A^2}{\cos ^2}\omega t$

$ = {1 \over 2}m{\omega ^2}{A^2}\left( {{{1 + \cos 2\omega t} \over 2}} \right)$ [$\because$ ${\cos ^2}\theta = {{1 + \cos 2\theta } \over 2}$]

$ = {1 \over 4}m{\omega ^2}{A^2}(1 + \cos 2\omega t)$

$\therefore$ New angular frequency, ${\omega _K} = 2\omega $

$\therefore$ Frequency of oscillation of kinetic energy = 2f [$\because$ $f = {\omega \over {2\pi }}$]

Let a be the acceleration of the lift.

Given, the ratio of time periods of pendulum while the lift is moving upwards and downwards,

T₁ : T₂ = 1 : 3

When lift is moving upwards, then total time period,

${T_1} = 2\pi \sqrt {{I \over {g + a}}} $........ (i)

When lift is moving downwards, then total time period,

${T_2} = 2\pi \sqrt {{I \over {g - a}}} $ ....... (ii)

Dividing Eq. (i) by Eq. (ii), we get

${{{T_1}} \over {{T_2}}} = \sqrt {{{g - a} \over {g + a}}} \Rightarrow {1 \over 3} = \sqrt {{{g - a} \over {g + a}}} $

or, ${{g - a} \over {g + a}} = {\left( {{1 \over 3}} \right)^2}$ or, ${{g - a} \over {g + a}} = {1 \over 9}$

$ \Rightarrow 9g - 9a = g + a$

$ \Rightarrow 9g - g = 9a + a$

$ \Rightarrow 10a = 8g$

$a = {{8g} \over {10}} = {{8 \times 10} \over {10}} = 8$ m/s²