 # A string is stretched between fixed points separated by 75 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is

### Frequency

The definition of frequency, we can understand that if a body is in periodic motion, it has undergone one cycle after passing through a series of events or positions and returning to its original state. Thus, frequency is a parameter that describes the rate of oscillation and vibration.

The relation between the frequency and the period is given by the equation:

f=1/T

Given

String is stretched between fixed point separated by the distance = 75 cm
Therefore, L = 75cm
The resonant frequencies of 420 Hz and 315 Hz.

Find out

We have to determine the lowest resonant frequency for this string

Solution

For the string fixed at both the ends, resonant frequency are given by f = nv/2L

Where,

n = smallest integer
v = velocity of sound
L = separation length of the string by which it is stretched.

Given that 315 Hz and 420 Hz are two consecutive resonant frequencies, let these be nth and (n + 1)th harmonics.

315 = nv/2L ——-(i)

420 = (n+1)v/2L——–(ii)

On dividing equa (i) by equation (ii) we get
n = 3

From equa (I), we get

lowest resonant frequency
f0 = v/2L

= 315/3

= 105 Hz

The lowest resonant frequency for this string is 105Hz

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