Physics > Simple Harmonic Motion > 11.0 Spring Block System
Simple Harmonic Motion
1.0 Types of Motion
2.0 Causes of Oscillation
3.0 Solution of the Equation of SHM
4.0 Kinematics of SHM equation i.e., $x = A\sin \left( {\omega t + \phi } \right)$
5.0 Relation between Simple Harmonic Motion and Uniform Circular Motion
6.0 Sign Convention of a Simple Harmonic Motion
7.0 How to Write the Simple Harmonic Motion Equation
8.0 Force and Energy in Simple Harmonic Motion
9.0 Basic Differential Equation of SHM
10.0 Method for Calculating Time Period of a Simple Harmonic Motion
10.1 Restoring Force or Torque Method
10.2 Energy Method
10.3 Important points
10.4 Concept of Pseudo Force
11.0 Spring Block System
12.0 Physical Pendulum
13.0 Vector Method of Combining Two or More Simple Harmonic Motions
14.0 Simple Harmonic Oscillation of a Fluid Column
11.1 Different Types of Spring Block Systems
10.2 Energy Method
10.3 Important points
10.4 Concept of Pseudo Force
(a). When the spring block system is vertical.
At equilibrium position,
$$k{x_0} = mg\quad ..\left( i \right)$$
Now, after displacing the block downward by a small distance $x$, we can write,
$$\begin{equation} \begin{aligned} k\left( {{x_0} + x} \right) - mg = - ma \\ ma = - kx \\ a = - \frac{k}{m}x\quad ..\left( {ii} \right) \\ a = - {\omega ^2}x\quad ..\left( {iii} \right) \\\end{aligned} \end{equation} $$
From, equation $(ii)$ & $(iii)$, we get,
$$\begin{equation} \begin{aligned} \omega = \sqrt {\frac{k}{m}} \quad ..\left( {iv} \right) \\ T = \frac{{2\pi }}{\omega } = 2\pi \sqrt {\frac{m}{k}} \\\end{aligned} \end{equation} $$
(b). Combination of Spring
- Parallel Combination:
All these combinations shown in Fig SHM 20 are for parallel combination of springs.
In all the cases, $${k_{eq}} = {k_1} + {k_2}$$ where ${k_{eq}}$ is 'equivalent spring constant’.
Proof: According to the FBD at displaced position,
$$\begin{equation} \begin{aligned} {k_{eq}}x = \left( {{k_1} + {k_2}} \right) \\ {k_{eq}} = {k_1} + {k_2} \\\end{aligned} \end{equation} $$ where ${k_{eq}}$ is 'equivalent spring constant’.
- Series Combination:
All these combinations shown in Fig SHM 22 are for series combination of springs.
In all the cases,
$$\frac{1}{{{k_{eq}}}} = \frac{1}{{{k_1}}} + \frac{1}{{{k_2}}}$$ where ${k_{eq}}$ is 'equivalent spring constant’.
Proof:
$${k_1}{x_1} - {k_2}{x_2} = {m_p}a$$ where $m_p$, is the mass of point $P$. Since point $P$ is massless, $m_p$$=0$.
$$\begin{equation} \begin{aligned} \therefore {k_1}{x_1} = {k_2}{x_2} \\ \Rightarrow {x_1} = \frac{{{k_2}{x_2}}}{{{k_1}}}\quad ..\left( i \right) \\\end{aligned} \end{equation} $$ Also, $${x_1} + {x_2} = x\quad ..\left( {ii} \right)$$
From equation $(i)$ & $(ii)$ we get, $$\begin{equation} \begin{aligned} \frac{{{k_2}{x_2}}}{{{k_1}}} + {x_2} = x \\ {x_2} = \frac{{{k_1}x}}{{\left( {{k_1} + {k_2}} \right)}}\quad ..\left( {iii} \right) \\\end{aligned} \end{equation} $$ From, FBD $${k_2}{x_2} = {k_{eq}}x\quad ..\left( {iv} \right)$$ From equation $(iii)$ & $(iv)$ we get,
$$\begin{equation} \begin{aligned} \frac{{{k_2}{k_1}x}}{{\left( {{k_1} + {k_2}} \right)}} = {k_{eq}}x \\ \frac{1}{{{k_{eq}}}} = \frac{1}{{{k_1}}} + \frac{1}{{{k_2}}} \\\end{aligned} \end{equation} $$ where ${k_{eq}}$ is 'equivalent spring constant’.
(c). Reduced Mass System
If two masses $m_1$ and $m_2$ are connected by spring and made to oscillate on horizontal surface. The time period is give by. $$T = 2\pi \sqrt {\frac{\mu }{k}} $$ where, $$\mu = \frac{{{m_1}{m_2}}}{{{m_1} + {m_2}}}$$ is known as reduced mass.
Proof:
The spring is elongated by $x$ distance.
Taking right direction as +ve and left direction as –ve.
Relative acceleration of block $1$ w.r.t. block $2$,
$$\begin{equation} \begin{aligned} \overrightarrow {{a_r}} = \overrightarrow {{a_1}} - \overrightarrow {{a_2}} \\ {a_r} = {a_1} - \left( { - {a_2}} \right) \\ {a_r} = {a_1} + {a_2} \\ - {\omega ^2}x = - \left( {\frac{k}{{{m_1}}} + \frac{k}{{{m_2}}}} \right) \\ \omega = \sqrt {\frac{{k\left( {{m_1} + {m_2}} \right)}}{{{m_1}{m_2}}}} \\ \omega = \sqrt {\frac{k}{\mu }} \\ \therefore T = 2\pi \sqrt {\frac{\mu }{k}} \quad {\text{where }}\mu = \frac{{\left( {{m_1} + {m_2}} \right)}}{{{m_1}{m_2}}} \\\end{aligned} \end{equation} $$
(d). Breaking of Spring
The force constant $k$ of a spring is inversely proportional to the length of the spring, i.e.
$$\begin{equation} \begin{aligned} k \propto \frac{1}{{{\text{length of the spring}}}} \\ \therefore kl = constant \\\end{aligned} \end{equation} $$
Suppose if a spring of length $l$ and spring constant $k$ is broken into two parts of length $\frac{l}{3}$ and $\frac{2l}{3}$ . Then find the springs constants of the respected length.
Since the spring is broken into two parts,
But,
$$\begin{equation} \begin{aligned} kl = constant \\ \therefore kl = {k_1}\frac{l}{3} = {k_2}\frac{{2l}}{3} \\ \Rightarrow {k_1} = 3k\quad {\text{and}}\quad {k_2} = \frac{{3k}}{2} \\\end{aligned} \end{equation} $$
(e). Spring Having Mass
Till now we have studied only about the massless spring.
Now, let us consider a block of mass ($m$) attached to a spring of mass ($m_s$). Then the time period is given as, $$T = 2\pi \sqrt {\frac{{m + \frac{{{m_s}}}{3}}}{k}} $$
Example 5. Find the time period of the arrangement as shown in the Fig SHM 27 (a) & (b).
Solution: (a). As observed form the figure all the spring are in the parallel combination.
Therefore, $${k_{eq}} = {k_1} + {k_2} + {k_3} + {k_4} + {k_5}\quad ..\left( i \right)$$ We know that the time period is given as, $$\begin{equation} \begin{aligned} T = 2\pi \sqrt {\frac{l}{{{k_{eq}}}}} \\ T = 2\pi \sqrt {\frac{l}{{{k_1} + {k_2} + {k_3} + {k_4} + {k_5}}}} \\\end{aligned} \end{equation} $$
(b). As observed from the figure, spring $1$ & $2$ and spring $5$ & $6$ are in series.
So, $$\begin{equation} \begin{aligned} \frac{1}{{{k_{12}}}} = \frac{1}{{{k_1}}} + \frac{1}{{{k_2}}} = \frac{1}{k} + \frac{1}{k} \\ {k_{12}} = \frac{k}{2} \\\end{aligned} \end{equation} $$ Similarly, $$\begin{equation} \begin{aligned} \frac{1}{{{k_{56}}}} = \frac{1}{{{k_5}}} + \frac{1}{{{k_6}}} = \frac{1}{k} + \frac{1}{{2k}} \\ {k_{56}} = \frac{{2k}}{3} \\\end{aligned} \end{equation} $$
Now the combination is reduced to the arrangement as shown in the Fig 27 (c).
Now, in this arrangement all the spring are in parallel combination.
Therefore,
$$\begin{equation} \begin{aligned} {k_{eq}} = {k_{12}} + {k_3} + {k_4} + {k_{56}} \\ {k_{eq}} = \frac{k}{2} + 2k + k + \frac{{2k}}{3} = \frac{{25k}}{6} \\\end{aligned} \end{equation} $$
Time period can be written as, $$T = 2\pi \sqrt {\frac{m}{{{k_{eq}}}}} = 2\pi \sqrt {\frac{{6m}}{{25k}}} $$
Example 6. A solid cylinder of mass m and radius R is attached to the massless spring as shown in the Fig SHM 28 (a) & (b). Find the period of oscillation in both the cases.
Note: The solid cylinder is rolling without slipping.
Solution: (a).
As shown in the Fig. SHM 28 c, solid cylinder is displaced by a distance $x$.
Since the spring is tied to the center of the cylinder, therefore the extension in spring will be also $x$.
From FBD, we write the energy of the system in the displaced position.
$$E{\text{ = Potential Energy of Spring + Translation Kinetic Energy + Rotational Kinetic Energy}}$$$$E = \frac{1}{2}k{x^2} + \frac{1}{2}m{v^2} + \frac{1}{2}I{\omega ^2}$$
For solid cylinder, $$I = \frac{1}{2}m{R^2}$$
As cylinder is rolling without slipping, $\omega = \frac{v}{R}$
$$\begin{equation} \begin{aligned} E = \frac{1}{2}k{x^2} + \frac{1}{2}m{v^2} + \frac{1}{2}\left( {\frac{{m{R^2}}}{2}} \right){\left( {\frac{v}{R}} \right)^2} \\ E = \frac{1}{2}k{x^2} + \frac{1}{2}m{v^2} + \frac{1}{4}m{v^2} \\ E = \frac{1}{2}k{x^2} + \frac{3}{4}m{v^2} \\\end{aligned} \end{equation} $$
Since $E=$ constant, therefore $\frac{{dE}}{{dt}} = 0$,
$$\begin{equation} \begin{aligned} \frac{{dE}}{{dt}} = \frac{1}{2}k\frac{{d{x^2}}}{{dt}} + \frac{3}{4}m\frac{{d{v^2}}}{{dt}} \\ 0 = kxv + \frac{3}{2}mva \\\end{aligned} \end{equation} $$
Since, $v \ne 0$,
$$\begin{equation} \begin{aligned} a = - \frac{{2kx}}{{3m}} \\ a = - {\omega ^2}x \\ \omega = \sqrt {\frac{{2k}}{{3m}}} \\ T = \frac{{2\pi }}{\omega } = 2\pi \sqrt {\frac{{3m}}{{2k}}} \\\end{aligned} \end{equation} $$
(b).
After displacing the cylinder by a distance $x$, the cylinder is also rotated by a small angle $\theta$.
Since the solid cylinder is rolling without slipping, therefore $x = R\theta $ and $v = \omega R$.
Extension in the spring $1$ is due $=$ linear displacement $+$ angular displacement i.e., $${x_1} = x + \frac{x}{2} = \frac{{3x}}{2}$$
Depression in the spring $2$ is due $=$ linear displacement $+$ angular displacement
$${x_2} = x + x = 2x$$
Now, we can write the equation for the energy of the system in the displaced position.
$$E{\text{ = Potential Energy }}\left( {{\text{Spring1+Spring2}}} \right){\text{ +Translational Kinetic Energy+Rotational Kinetic Energy}}$$
$$\begin{equation} \begin{aligned} E = \frac{1}{2}k{\left( {\frac{{3x}}{2}} \right)^2} + \frac{1}{2}2k{\left( {2x} \right)^2} + \frac{1}{2}m{v^2} + \frac{1}{2}I{\omega ^2} \\ E = \frac{{41}}{8}k{x^2} + \frac{1}{2}m{v^2} + \frac{1}{2}I{\omega ^2} \\\end{aligned} \end{equation} $$
For solid cylinder, $I = \frac{1}{2}m{R^2}$
Also, solid cylinder is rolling without slipping, $\omega = \frac{v}{R}$
$$\begin{equation} \begin{aligned} E = \frac{{41}}{8}k{x^2} + \frac{1}{2}m{v^2} + \frac{1}{2}\left( {\frac{{m{R^2}}}{2}} \right){\left( {\frac{v}{R}} \right)^2} \\ E = \frac{{41}}{8}k{x^2} + \frac{1}{2}m{v^2} + \frac{1}{4}m{v^2} \\ E = \frac{{41}}{8}k{x^2} + \frac{3}{4}m{v^2} \\\end{aligned} \end{equation} $$
Since energy is constant, therefore, $\frac{{dE}}{{dt}} = 0$
$$\begin{equation} \begin{aligned} \frac{{dE}}{{dt}} = \frac{{41}}{8}k\left( {\frac{{d{x^2}}}{{dt}}} \right) + \frac{3}{4}m\left( {\frac{{d{v^2}}}{{dt}}} \right) \\ 0 = \frac{{41}}{4}kxv + \frac{3}{2}mva \\\end{aligned} \end{equation} $$
Since, $v \ne 0$,
$$\begin{equation} \begin{aligned} a = - \frac{{41kx}}{{6m}} \\ \because a = - {\omega ^2}x \\ \Rightarrow \omega = \sqrt {\frac{{41k}}{{6m}}} \\ \therefore T = \frac{{2\pi }}{\omega } = 2\pi \sqrt {\frac{{6m}}{{41k}}} \\\end{aligned} \end{equation} $$
Example 7. Find the time period of the arrangement as shown in the Fig SHM 29 (a) & (b).
Solution: (a).
At equilibrium position,
Since the spring $1$ is displaced by $x_1$ and spring $2$ by $x_2$.
Therefore the block is displaced by $$x = \left( {2{x_1} + 2{x_2}} \right)\quad ..\left( i \right)$$
From the Fig SHM 30, we can write,
$$\begin{equation} \begin{aligned} 2T = {k_1}{x_1}\left( {{\text{Fig SHM }}30b} \right) \\ {x_1} = \frac{{2T}}{{{k_1}}}\quad ..\left( {ii} \right) \\ 2T = {k_2}{x_2}\left( {{\text{Fig SHM }}30c} \right) \\ {x_1} = \frac{{2T}}{{{k_1}}}\quad ..\left( {iii} \right) \\\end{aligned} \end{equation} $$
Also, $$T = mg\left( {{\text{Fig SHM }}30d} \right)\quad ..\left( {iv} \right)$$
From equation $(i)$, $(ii)$, $(iii)$ & $(iv)$ we get,
$$x = 4mg\left( {\frac{1}{{{k_1}}} + \frac{1}{{{k_2}}}} \right)\quad ..\left( v \right)$$
The above arrangement is now reduced to the arrangement as shown in Fig. SHM 30 e.
From Fig. SHM 30 e, we get, $${k_{eq}}x = mg\quad ..\left( {vi} \right)$$
From equation $(v)$ & $(vi)$, we get
$$\begin{equation} \begin{aligned} {k_{eq}}4mg\left( {\frac{1}{{{k_1}}} + \frac{1}{{{k_2}}}} \right) = mg \\ {k_{eq}} = \frac{{{k_1}{k_2}}}{{4\left( {{k_1} + {k_2}} \right)}}\quad ..\left( {vii} \right) \\\end{aligned} \end{equation} $$
The time period of the vertical spring arrangement is given by,
$$\begin{equation} \begin{aligned} T = 2\pi \sqrt {\frac{m}{{{k_{eq}}}}} \\ T = 2\pi \sqrt {\frac{{4m\left( {{k_1} + {k_2}} \right)}}{{{k_1}{k_2}}}} \\\end{aligned} \end{equation} $$
(b).
At equilibrium position,
Since the spring $1$ is displaced by $x_1$ and spring $2$ by $x_2$.
Therefore the block is displaced by $$x = \left( {2{x_1} + {x_2}} \right)\quad ..\left( i \right)$$
From the Fig SHM 30, we can write,
$$\begin{equation} \begin{aligned} 2T = {k_1}{x_1}\left( {{\text{Fig SHM }}31b} \right) \\ {x_1} = \frac{{2T}}{{{k_1}}}\quad ..\left( {ii} \right) \\ T = {k_2}{x_2}\left( {{\text{Fig SHM }}31c} \right) \\ {x_1} = \frac{T}{{{k_1}}}\quad .\left( {iii} \right) \\\end{aligned} \end{equation} $$
From equation $(i)$, $(ii)$, $(iii)$ & $(iv)$ we get,
$$x = mg\left( {\frac{4}{{{k_1}}} + \frac{1}{{{k_2}}}} \right)\quad ..\left( v \right)$$
The above arrangement is now reduced to the arrangement as shown in Fig. SHM 31 e.
From Fig. SHM 31 e, we get $${k_{eq}}x = mg..\left( {vi} \right)$$
From equation $(v)$ & $(vi)$, we get
$${k_{eq}}mg\left( {\frac{4}{{{k_1}}} + \frac{1}{{{k_2}}}} \right) = mg$$$${k_{eq}} = \frac{{{k_1}{k_2}}}{{\left( {{k_1} + 4{k_2}} \right)}}\quad ..\left( {vii} \right)$$
The time period of the vertical spring arrangement is given by, $$\begin{equation} \begin{aligned} T = 2\pi \sqrt {\frac{m}{{{k_{eq}}}}} \\ \therefore T = 2\pi \sqrt {\frac{{m\left( {{k_1} + 4{k_2}} \right)}}{{{k_1}{k_2}}}} \\\end{aligned} \end{equation} $$
