Question

Define S.H.M. What does phase of oscillation signify? A horizontal spring block system of force constant $K$ and mass $M$ executing in SHM with amplitude $A$. When the block is passing through its mean position an object of mass $m$ is put on it and two move together. Find a new amplitude and frequency of oscillation?

Answer 1

In order to calculate the new amplitude and the new frequency, we need to understand the concept that when a body is present at its mean position while oscillating, ten at that point it has the maximum kinetic energy and this kinetic energy is equal to the total energy. Also, in order to calculate the new frequency, we will calculate the time period as the frequency is the reciprocal of the time period.

Complete step by step answer:

The motion of any object which moves to and fro about its mean position along a straight line is known as simple harmonic motion. A simple pendulum undergoes simple harmonic motion. It swings to and fro about its mean position and thus, we can say that the pendulum undergoes simple harmonic motion. Phase basically is an angular term which is used to represent the situation of a particle in SHM at a certain point of time. Phase of oscillation is also defined as a state of oscillation of particles performing simple harmonic motion. When the system of mass is passing by the mean position, then the kinetic energy will be maximum and it will be equal to the total energy. So,
Total energy $=$ Kinetic energy

On putting the expression of kinetic energy and total energy, we get,
$\dfrac{1}{2} \times K \times {A^2} = \dfrac{1}{2} \times M \times {v^2}$
$K \times {A^2} = M \times {v^2}$
On further simplifying,
$v = \sqrt {\dfrac{{K{A^2}}}{M}}$
$\Rightarrow v = \sqrt {\dfrac{K}{M}} A.....(1)$
According to the law of conservation of linear momentum,
Final momentum $=$ Initial momentum
$(M + m){v^1} = Mv$
On further solving, we get,
${v^1} = \dfrac{{Mv}}{{M + m}}.......(2)$
When the two blocks form a new system, then also the kinetic energy at the mean position will be equal to the total energy. So,
Total energy $=$ Kinetic energy

On putting the expression of kinetic energy and total energy, we get,
$\dfrac{1}{2} \times K \times {({A^1})^2} = \dfrac{1}{2} \times (M + m) \times {({v^1})^2}$
On putting the value of ${v^1}$ from equation (2), we get,
$\dfrac{1}{2} \times K \times {({A^1})^2} = \dfrac{1}{2} \times (M + m) \times {(v)^2} \times \left( {\dfrac{M}{{M + m}}} \right)\left( {\dfrac{M}{{M + m}}} \right)$
On simplifying the above equation, we get,
$K \times {({A^1})^2} = {(v)^2} \times \left( {\dfrac{M}{{M + m}}} \right) \times M$
On taking ${A^2}$ on one side and all other terms on the other side, we get,
${\left( {{A^1}} \right)^2} = \dfrac{{{M^2}{v^2}}}{{K(M + m)}}$
On taking square root on both the sides, we get,
${A^1} = \dfrac{{Mv}}{{\sqrt {M + m} \sqrt K }}$

Now, we need to calculate the new frequency. We know that there is an inverse relation between time period and frequency, i.e.
$\nu = \dfrac{1}{T}$
So, first we need to calculate the time period.
$T = 2\pi \sqrt {\dfrac{I}{K}}$
$\Rightarrow T = 2\pi \sqrt {\dfrac{{M + m}}{K}}$
On taking the reciprocal of the above equation, we get the new frequency,
$\nu = \dfrac{1}{{2\pi \sqrt {\dfrac{{M + m}}{K}} }}$
$\therefore \nu = \dfrac{1}{{2\pi }}\sqrt {\dfrac{K}{{M + m}}}$

Therefore, the new amplitude is ${A^1} = \dfrac{{Mv}}{{\sqrt {M + m} \sqrt K }}$ and the new frequency is $\nu = \dfrac{1}{{2\pi }}\sqrt {\dfrac{K}{{M + m}}}$.

Note: When a pendulum swings and performs simple harmonic motion, then a continuous transformation of energy takes place. When the bob is at the extreme position, then there is maximum potential energy and minimum kinetic energy. Whereas when the bob is at the mean position, then the pendulum has the maximum kinetic energy and minimum potential energy.