Oscillations Of A Spring-mass System MCQ - Practice Questions with Answers

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, the -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 the Force method.

Let $x_0$ be the deformation in the spring in equilibrium. Then $k{x}_0=\mathrm{mg}$. 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 figure below.

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

2. Oscillation of spring in the horizontal  plane

For the above figure, Using the force method we get a Time period of spring as

$$T=2\pi \sqrt{\frac{m}{k}}$$
 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 of a spring pendulum 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 the spring is halved then its force constant becomes double.

5. When a spring of length $I$ is cut in two pieces of length $l_1$ and $I_2$ such that $l_1=n{l}_2$

So using

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

$$\text{ similarly }{l}_1=n{l}_2\Rightarrow l_1=\frac{l\ast n}{(n+1)}$$

If the constant of a spring is $k$ then
$u\mathrm{sing}kl=\mathrm{constant}$

$$\text{ i.e }{k}_1{l}_1=k_2{l}_2=kl$$

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

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

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

2. The parallel combination of spring

If 2 springs of different force constants 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 $1\mathrm{\&}2$ respectively.
Similarly, If $n$ springs of different force constants are connected in parallel having force constants $k_1,k_2,k_3\dots \dots$ respectively
then $K_{eq}=K_1+K_2+K_3\dots$
If all the n springs have the same spring constant as $K_1$ then $K_{eq}=n{K}_1$