Question

The energy E of a particle oscillating in SHM depends on the mass m of the particle, frequency n, and amplitude a of oscillation. Show dimensionally that $E\propto kmn^{2}a^{2}$.

Answer 1

Hint: Any physical quantity's dimension can be expressed as a function of the symbols (or powers of symbols) that represent the base quantities. To answer this question, first, find the dimension of the left-hand side term which is energy E. Then, proceed further by getting the dimension of each term on the right-hand side. This will help in proving the given relation dimensionally.

Complete step by step solution:
Dimensional analysis is the process of determining the dimensions of physical quantities to examine relationships between them. All quantities in the universe can be stated as a function of the basic dimensions, which are independent of numerical multiples and constants.

In this equation, we need to show dimensionally that energy E of a particle oscillating in SHM depends on the mass m of the particle, frequency n, and amplitude a of oscillation. Simple harmonic motion is a particular sort of periodic motion in mechanics and physics in which the restoring force on the moving item is directly proportional to the magnitude of the displacement and operates in the direction of the object's equilibrium position.

The dimensional formula of energy can be written as:
$E=\left [ M^{1}L^{2}T^{-2}\right ]$ (Since energy is equivalent to work done which is the product of force and displacement)
Now, the dimension formula of mass is: $\left [ M^{1}\right ]$
Similarly, for frequency, it is: $\left [ T^{-1}\right ]$(where frequency is the inverse of time period T)
Also, for amplitude: $\left [ L^{1}\right ]$
Let us consider the power of the dimensional formula of mass, frequency, and amplitude to be x, y, and z.Thus,
$\left [ E \right ]=k\left [ M^{1} \right ]^{x}\left ( T^{-1} \right )^{y}\left [ L^{1} \right ]^{z}$ where, k is the constant.

Substitute the dimensional formula of energy in the above equation,
$\left [ M^{1}L^{2}T^{-2}\right ]=\left [ M^{1} \right ]^{x}\left ( T^{-1} \right )^{y}\left [ L^{1} \right ]^{z}$
Comparing LHS and RHS and finding the value of x, y, and z, we get
$x=1$
$y=2$
$z=2$
So, the required equation is
$E=kmn^{2}a^{2}$

Hence, $E\propto kmn^{2}a^{2}$ (where, k is the proportionality constant).

Note: We need to ensure that the equation is dimensionally correct. To determine the dimensional correctness of a given physical relationship, we must first determine whether the dimensions on both sides of each term are the same or not; if they are not, the equation is dimensionally incorrect; if the dimensions on both sides of each term of the given equation are the same, the equation is dimensionally correct.