Question

The frequency of vibrations f of a mass m suspended from a spring of spring constant K is given by a relation of type $f = c{m^x}{k^y}$,where c is a dimensionless constant. The values of x and y are:

Answer 1

Hint: Use the idea that frequency is the inverse of time period and that the dimension of time period is$[T]$ in this question. So, using the dimension of frequency values of x and y can be find out.

Complete answer:
Physics defines frequency as the quantity of waves passing through a fixed location in relation to a unit of time. The quantity of cycles or vibrations that a body experiences in one unit of time with periodic motion is another way to describe frequency. The body can move in cycles or vibrations that go through a succession of states or postures before returning to its initial position as part of its periodic motion. This is comparable to the simple harmonic motion or angular velocity.

The high-frequency waves were calculated using an electrical equipment known as a frequency counter. The outcome is displayed in Hertz via the frequency counter. The Heterodyne method can be used to determine a wave's frequency that is higher than that of a frequency counter.

We can say that frequency (f) is inverse of time period (T).
$ \Rightarrow f = \dfrac{1}{T}$
$ \Rightarrow f = [{T^{ - 1}}]$
Since the spring force is given by $F = kx$

Dimension of spring constant, k = dimension of F/dimension of x
$ \Rightarrow k = \dfrac{{[ML{T^{ - 2}}]}}{{[L]}} = [M{T^{ - 2}}]$

Now putting their dimensions value in relation $f = c{m^x}{k^y}$we get,
${T^{ - 1}} = c[{M^x}]{[M{T^{ - 2}}]^y}$
$ \Rightarrow {T^{ - 1}} = c[{M^{x + y}}][{T^{ - 2y}}]$

On comparing both sides, we get
$x + y = 0$and $ - 2y = - 1$

So we can say that $x = - \dfrac{1}{2}$and $y = \dfrac{1}{2}$

Note: The number of waves that pass a fixed place in a unit of time is referred to as frequency. It may also be described as how many cycles or vibrations a body experiences in a unit of time when it engages in a periodic motion that repeats itself after a set amount of time.