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

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• Oscillations in combination of springs is considered one the most difficult concept.

• Spring System is considered one of the most asked concept.

• 33 Questions around this concept.

## Solve by difficulty

A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period T. If the mass is increased by m, the time period becomes 5T/3. Then the ratio of $\dpi{100} m/M$ is :

If a spring has time period T , and is cut into $\dpi{100} n$ equal parts, then the time period of each part will be

For a simple harmonic motion in a mass spring system shown, the surface is frictionless. When the mass of the block is $1 \mathrm{~kg}$, the angular frequency is $\omega_{1}$. When the mass block is $2 \mathrm{~kg}$ the angular frequency is $\omega_{2}$. The ratio $\omega_{2} / \omega_{1}$ is

A particle at the end of a spring executes simple harmonic motion with a period t1 , while the corresponding period for another spring is t2 . If the period of oscillation with the two springs in series is T, then

A 1 kg block attached to a spring vibrates with a frequency of 1 Hz on a frictionless horizontal table.  Two springs identical to the original spring are attached in parallel to an 8 kg block placed on the same table. So, the frequency of vibration of the 8 kg block is :

Two springs, of force constants and are connected to a mass as shown. The frequency of oscillation of the mass is .      If both and are made four times their original values, the frequency of oscillation becomes

## Concepts Covered - 2

Spring System

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.

using  $F=-\left[k\left(x+x_{0}\right)-m g\right]$ and $k x_{0}=m g .$

we get $F=-kx$

comparing it with the equation of SHM i.e $F=-m\omega ^2x$

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

• 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.

$\text { As } \quad k \propto \frac{1}{\text { Extension }} \propto \frac{1}{\text { Length of spring }(l)}$

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 l is cut in two pieces of length l1 and l2 such that  $l_1=nl_2$

So using

$\dpi{120} l_1+l_2=l \\ \begin{array}{l} nl_{2}+l_{2}=l \\ (n+1) l_{2}=l \Rightarrow l_2=\frac{l}{n+1}\\ \\ \text{similarly} \ \l_{1}=nl_2 \Rightarrow l_1=\frac{l*n}{(n+1)} \end{array}$

If the constant of a spring is k then

using $kl=constant$

i.e $\dpi{120} k_1l_1=k_2l_2=kl$

we get

$\begin{array}{l}{\text { Spring constant of first part } k_{1}=\frac{k(n+1)}{n}} \\ {\text { Spring constant of second part } k_{2}=(n+1) k}\end{array}$

$\text { and ratio of spring constant } \frac{k_{1}}{k_{2}}=\frac{1}{n}$

6.  If the spring has a mass M and mass m is suspended from it, then its effective mass is given by $m_{e f f}=m+\frac{M}{3}$

and $T=2 \pi \sqrt{\frac{m_{e f f}}{k}}$

Oscillations in combination of springs

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 & 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 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 & 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

then $K_{eq}=K_{1}+K_{2}+K_{3}....$

If all the n spring have the same spring constant as $K_{1}$ then   $K_{eq}=nK_{1}$

## Study it with Videos

Spring System
Oscillations in combination of springs

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## Books

### Reference Books

#### Spring System

Physics Part II Textbook for Class XI

Page No. : 352

Line : 53