Oscillations Of A Spring-mass System - Practice Questions & MCQ

Spring Force:-

• Spring force is also called restoring force.

• F= -Kx 

           where k is the spring constant and its unit is N/m and x is net elongation or compression in the spring.

           Spring constant (k) is a measure of stiffness or softness of the spring

• Here -ve sign is because the force exerted by the spring is always in the opposite direction to the displacement. 

The time period of the Spring mass system-

1. Oscillation of a spring in a verticle plane-

Finding Time period of spring using Force method.

Let $x_0$ be the deformation in the spring in equilibrium. Then $k{x}_0=mg$. When the block is further displaced by x , the net restoring force is given by $F=-[k(x+x_0)-mg]$ as shown in the below figure.e.

using $F=-{[k(x+x_0)-mg]}_{\text{and }}k{x}_0=mg$.
we get $F=-kx$
comparing it with the equation of SHM i.e $F=-m{\omega }^{2}x$
we get

${\omega }^2=\frac{k}{m}\Rightarrow T=2\pi \sqrt{\frac{m}{k}}$

similarly

$\text{ Frequency }=n=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$

2.Oscillation of spring in the horizontal plane

For the above figure, Using force method we get a Time period of spring as $T=2\pi \sqrt{\frac{m}{k}}$ and Frequency $=n=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$
- Key points
1. The time period of a spring-mass system depends on the mass suspended

$T\propto \sqrt{m}\text{ or }n\propto \frac{1}{\sqrt{m}}$

2. The time period of a spring-mass system depends on the force constant k of the spring

$T\propto \frac{1}{\sqrt{k}}\text{ or }n\propto \sqrt{k}$

3. The time period of a spring-mass system is independent of acceleration due to gravity.
4. The spring constant $k$ is inversely proportional to the spring length.

As $k\propto \frac{1}{\text{ Extension }}\propto \frac{1}{\text{ Length of spring (l) }}$
i.e $kl=$ constant

That means if the length of spring is halved then its force constant becomes double.
5. When a spring of length I is cut in two pieces of length ${\mathrm{l}}_1$ and ${\mathrm{l}}_2$ such that $l_1=n{l}_2$

     So using

$\begin{array}{rl}& l_1+l_2=l\\ & n{l}_2+l_2=l\\ & (n+1)l_2=l\Rightarrow l_2=\frac{l}{n+1}\end{array}$

similarly $l_1=n{l}_2\Rightarrow l_1=\frac{l\ast n}{(n+1)}$
If the constant of a spring is k then

$\text{ using }kl=\text{ constant }$

i.e $k_1{l}_1=k_2{l}_2=kl$
we get
Spring constant of first part $k_1=\frac{k(n+1)}{n}$
Spring constant of second part $k_2=(n+1)k$
and ratio of spring constant $\frac{k_1}{k_2}=\frac{1}{n}$
6. If the spring has a mass $M$ and mass $m$ is suspended from it, then its effective mass is given by

$\begin{array}{rl}{m}_{eff}& =m+\frac{M}{3}\\ & T=2\pi \sqrt{\frac{m_{eff}}{k}}\end{array}$
 

1. Series combination of spring

If 2 springs of different force constant are connected in series as shown in the below figure

then k=equivalent force constant is given by 

$\frac{1}{K_{eq}}=\frac{1}{K}=\frac{1}{K_1}+\frac{1}{K_2}$

Where $K_1$ and $K_2$ are spring constants of spring $1\mathrm{\&}2$ respectively.
Similarly, If n springs of different force constant are connected in series having force constant $k_1,k_2,k_3\dots \dots$ respectively
then

$\frac{1}{k_{eff}}=\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}+\dots \dots \dots$

If all the n spring have the same spring constant as $K_1$ then $\frac{1}{k_{eff}}=\frac{n}{k_1}$

2.The parallel combination of spring

If 2 springs of different force constant are connected in parallel as shown in the below figure

then $\mathrm{k}=$ equivalent force constant is given by

$K_{eq}=K=K_1+K_2$

where $K_1$ and $K_2$ are spring constants of spring $1\mathrm{\&}2$ respectively.
Similarly, If $n$ springs of different force constant are connected in parallel having force constant $k_1,k_2,k_3\dots \dots$ respectively

$\text{ then }{K}_{eq}=K_1+K_2+K_3\dots$

If all the n spring have the same spring constant as $K_1$ then $K_{eq}=n{K}_1$