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 $k x_0=m g$. When the block is further displaced by x , the net restoring force is given by $F=-\left[k\left(x+x_0\right)-m g\right]$ as shown in the below figure.e.

using $F=-\left[k\left(x+x_0\right)-m g\right]_{\text {and }} k x_0=m g$.
we get $F=-k x$
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} \quad \text { or } \quad 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}} \quad \text { or } \quad 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 $\quad k \propto \frac{1}{\text { Extension }} \propto \frac{1}{\text { Length of spring (l) }}$
i.e $k l=$ 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{aligned}
& \quad l_1+l_2=l \\
& n l_2+l_2=l \\
& (n+1) l_2=l \Rightarrow l_2=\frac{l}{n+1}
\end{aligned}
$

similarly $l_1=n l_2 \Rightarrow l_1=\frac{l * n}{(n+1)}$
If the constant of a spring is k then

$
\text { using } k l=\text { constant }
$

i.e $k_1 l_1=k_2 l_2=k l$
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{aligned}
m_{e f f} & =m+\frac{M}{3} \\
& T=2 \pi \sqrt{\frac{m_{e f f}}{k}}
\end{aligned}
$
 

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_{e q}}=\frac{1}{K}=\frac{1}{K_1}+\frac{1}{K_2}
$

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

$
\frac{1}{k_{e f f}}=\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}+\ldots \ldots \ldots
$

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

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_{e q}=K=K_1+K_2
$

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

$
\text { then } K_{e q}=K_1+K_2+K_3 \ldots
$

If all the n spring have the same spring constant as $K_1$ then $K_{e q}=n K_1$