Question

A $1\;kg$ block is executing simple harmonic motion of amplitude over a horizontal smooth surface under a restoring force of a spring of spring constant $100\;N{m^{ - 1}}$. A block of mass $3\;kg$ is gently placed on it at the instant it passes through the mean position. Assuming that the two blocks move together, find the frequency and amplitude of the motion.

Answer 1

The spring undergoes oscillatory motion if it is stretched and tries to go back to its original position due to the restoring force that is applied on it. Since the whole system will be in motion the kinetic energy of the boxes need to be determined and the potential energy of the boxes also need to be found once the body reaches its mean position and the law of conservation of energy is applied.

Complete step by step answer:
The block is said to be executing simple harmonic motion. A body executing simple harmonic motion tends to move to and fro about a mean position which is its original position under the action of a restoring force that is directly proportional to its displacement. This happens due to the attached block which is considered as the load to the spring which is responsible for the force applied on it and the spring’s corresponding extension.

Knowing this concept, we can say that when the spring undergoes simple harmonic motion then there will be a displacement from the block’s mean position due to the oscillatory motion. The maximum displacement about the mean position is called the amplitude of the block. The frequency is the number of oscillations taken per unit time by the block to return to its mean position. We are asked to find the amplitude and frequency of the system, that is, when the second block is also placed above the first block.The first step is to extract the data given in the question.
Given, ${m_1} = 1\;kg$
$\Rightarrow {m_2} = 3\;kg$
$\Rightarrow k = 100\;N{m^{ - 1}}$
$\Rightarrow A = 0.1\,m$
where, ${m_1}$ and ${m_2}$ are the masses of the two blocks respectively and $k$ is given to be the spring constant. The amplitude of the harmonic motion executed by the first block is given.

The energy of a harmonic oscillator (the spring in this case) is said to be partly kinetic and partly potential. This means that the body poses both kinetic and potential energy and hence we need to find them out. First we take only the block with mass $1\;kg$, that is, the first block into consideration. The kinetic energy to be determined is dependent upon the velocity of motion which is in-turn dependent on the angular frequency. We know that the angular frequency is given by the formula:
${\omega ^2} = \dfrac{k}{m}$
$\Rightarrow \omega = \sqrt {\dfrac{k}{{{m_1}}}}$

By substituting the values of mass of first block and the spring constant we get $\omega$ to be:
$\omega = \sqrt {\dfrac{{100}}{1}}$
$\Rightarrow \omega = 10{\operatorname{s} ^{ - 1}}$
Now that we have determined the value of $\omega$ we can construct the equation for velocity and find its value. The velocity of a particle in simple harmonic motion is given by the equation:
$v = \omega \sqrt {{A^2} - {x^2}}$
But the block at mean position or its original position will have no displacement and hence $x = 0$. Thus the equation becomes:
$v = \omega A$

The amplitude of the first block was already given to us. By substituting the values in the above equation we get:
$v = 10 \times 0.1$
$\Rightarrow 1m{\operatorname{s} ^{ - 1}}$
This will be the velocity when the second block is not taken into consideration, that is, the second block is not placed over the first one.
Next, we come to the frequency of the system. The frequency of oscillation is given by the general formula:
$f = \dfrac{1}{T}$
where, $T$ is the time period of the oscillations
This time period for a body exhibiting simple harmonic motion is given by the formula:
$T = 2\pi \sqrt {\dfrac{m}{k}}$
Hence by substituting this into the frequency equation we get:
$f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}}$ -----($1$)

Now, we consider that the second block is also placed on top of the first block. These two blocks placed together will be considered as one system as it will move together as a single unit as given in the question. Hence the total mass of the system will be the sum of their respective masses.
Hence, $m = 3 + 1$
$\Rightarrow m = 4\;kg$
Therefore, by substituting the values into the equation ($1$) we get:
$f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{100}}{4}}$
$\Rightarrow \dfrac{1}{{2\pi }} \times \dfrac{{10}}{4}$
$\Rightarrow f = \dfrac{5}{{2\pi }}{\operatorname{s} ^{ - 1}}$
Hence the frequency of motion when the two blocks move together is $\dfrac{5}{{2\pi }}{\operatorname{s} ^{ - 1}}$.

Now the concept of conservation of energy or linear momentum comes into play. Initially there was only one block with mass $1\;kg$ but now we must consider the two blocks as a whole system and the velocity of this system needs to be found out which is the final momentum of the system. By the law of conservation of momentum we know that the initial momentum is equivalent to the final momentum of the system in-order to conserve energy. This is given by the equation:
${m_1}{v_1} = {m_2}{v_2}$
Here ${m_1}$ will be the mass of the first block only and its corresponding velocity ${v_1}$. But ${m_2}$ is the mass of the whole system so ${m_2}$ will be equal to the total mass of the two blocks that is $4\;kg$ and ${v_2}$ is the corresponding velocity of the whole system. Hence by substituting the values which were previously obtained we get:
$1 \times 1 = 4 \times {v_2}$
$\Rightarrow {v_2} = \dfrac{1}{4}m{s^{ - 1}}$

Hence we can now finally determine the kinetic energy of the system. This is calculated at the mean position of the blocks which will be its maximum value as there is no energy lost due to motion (displacement is zero). A point to be noted is that the potential energy at the mean position is zero since it is dependent on the displacement of the body. The kinetic energy is hence given by the equation:
${K_{\max }} = \dfrac{1}{2} \times {m_2}v_2^2$
$\Rightarrow K = \dfrac{1}{2} \times 4 \times {\left( {\dfrac{1}{4}} \right)^2}$
$\Rightarrow K = \dfrac{1}{8}J$
Next, we determine the potential energy of the system. This energy is the energy stored by the system when the displacement of the blocks are at its maximum which means that the displacement becomes the amplitude in accordance with its definition.

Hence, potential energy is given by the equation:
$P.E = \dfrac{1}{2} \times kA{}^2$
By the law of conservation of energy we can say that the kinetic energy of this system is equivalent to its potential energy. This is because the kinetic energy of the system which was the total energy of the system initially at the mean position is entirely converted into potential energy .Hence, by equating them we get:
$\Rightarrow K.E = P.E$
We now substitute the value of kinetic energy that was determined in the equation.
$\Rightarrow \dfrac{1}{8} = \dfrac{1}{2} \times 100 \times {A^2}$
Therefore we get the value for the amplitude if we solve out the above equation.
$\Rightarrow {A^2} = \dfrac{2}{{8 \times 100}}$
$\Rightarrow {A^2} = \dfrac{1}{{400}}$
$\Rightarrow A = \sqrt {\dfrac{1}{{400}}}$
$\Rightarrow A = \dfrac{1}{{20}}m$
$\therefore A = 0.05\;m$

Hence when the two blocks move together the frequency of motion is $\dfrac{5}{{2\pi }}{\operatorname{s} ^{ - 1}}$ and the amplitude is $0.05\;m$.

Additional information: When a body is displaced from its mean position there is some amount of work done which is stored in the form of potential energy and when body is released it moves back to its equilibrium position acquiring kinetic energy.

Hence a system obeying simple harmonic motion has both kinetic and potential energy. The motion of the particle in simple harmonic motion is periodic. A property of simple harmonic motion is said to be expressed in terms of single harmonic functions or sine and cosine terms.

Note: The force on a spring is a conservational force and hence the above problem uses the concept of law of conservation of energy where the kinetic energy is zero when the system's potential energy is maximum and the potential energy is zero when the kinetic energy is maximum. This simply means that at the equilibrium position the total energy of the system is solely kinetic and at the extreme position it is purely potential.