Question

If we add load to the hanger of a sonometer. The fundamental frequency becomes three times of the initial value. The initial load in the hangar was about?
(A) $4Kg$
(B) $2Kg$
(C) $1Kg$
(D) $0.5Kg$

Answer 1

Hint: We need to use the equation to find out the velocity of the wave and then put the value of this obtained velocity in the expression of frequency. The mass of the later frequency is eight more than the earlier one. Using the relation between the two frequencies, we can determine the value of initial load.

Complete Step-By-Step Solution:
We know, in the formula for velocity, it is written as:
$V = \sqrt {\dfrac{{TL}}{m}} $
Where,
$V = $ Velocity
$m = $ Mass of the string
$T = $ Tension
$L = $ Length of the string
We know, the formula for fundamental frequency is written as:
$f = \dfrac{V}{{2L}}$
Where,
\[f = \] Fundamental Frequency
$V = $ Velocity
$L = $ Length of the string
We know, when a body is hung, there exists a tension.
Tension is nothing but the drawing force that appears when an object is hung on a string.
The value of tension can be calculated by multiplying mass of the object with the acceleration due to gravity.
Let $M$ be the mass of the object hanging from the string
Therefore, the tension of the string is:
$T = Mg$
$T = $ Tension of the string
Putting the value of tension in the equation of velocity, we get:
$V = \sqrt {\dfrac{{MgL}}{m}} $
Putting this value of velocity in the equation of fundamental frequency, we get:
${f_1} = \dfrac{1}{{2L}}\sqrt {\dfrac{{MgL}}{m}} $
Now, in the second case, mass of \[8Kg\]is added, therefore, the mass becomes $ = (M + 8)Kg$
Therefore, equation of the fundamental frequency becomes:
${f_2} = \dfrac{1}{{2L}}\sqrt {\dfrac{{(M + 8)gL}}{m}} $
Now, in the question, it is given that
${f_2} = 3{f_1}$
Thus, we can write expression as:
$\dfrac{3}{{2L}}\sqrt {\dfrac{{MgL}}{m}} = \dfrac{1}{{2L}}\sqrt {\dfrac{{(M + 8)gL}}{m}} $
On cancelling the common terms and solving the equation, we obtain:
$M = 1Kg$
This is the initial mass of the object.

Hence, option (C) is correct.

Note:
Frequency of the object is directly proportional to its mass. Therefore, when mass of the body is increased, its frequency increases and similarly vice versa. As frequency increases, the pitch of the sound also increases. Pitch implies the highness or lowness of the sound.