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Question: The frequency of vibration \( (v) \) of a string may depend upon length \( (l) \) of the string, ten...

The frequency of vibration (v)(v) of a string may depend upon length (l)(l) of the string, tension (T)(T) in the string and mass per unit length (m)(m) of the string. Use the method of dimensions to determine the formula for frequency vv.

Explanation

Solution

Method of dimensions makes use of the fact that every physical quantity can be written in the form of combinations of various base units (or quantities). This method is used to identify the units of a new quantity when we know its relation with other parameters.

Complete step by step answer:

In this question, we are told that the frequency of vibration of a string depends on: length, tension and mass per unit length of the string. We are aware of the fundamental units of these three parameters. So, let us write them down in terms of mass, length and time:

v=[M0L0T1]v = [{M^0}{L^0}{T^{ - 1}}]

l=[M0L1T0]l = [{M^0}{L^1}{T^0}]

m=[M1L1T0]m = [{M^1}{L^{ - 1}}{T^0}]

T=[M1L1T2]T = [{M^1}{L^1}{T^{ - 2}}] ...............(As tension has the same units as force)

Now, let us assume the frequency directly depends on these quantities as:

vlambTc\Rightarrow v \propto {l^a}{m^b}{T^c}

Where aa , bb and cc are arbitrary constants whose values we will try to determine. Upon removing the proportionality sign, we get:

v=klambTc\Rightarrow v = k{l^a}{m^b}{T^c}

Where kk is the proportionality constant. We know input all he dimensions and then, use the method of comparison to find the values of aa , bb and cc :

[M0L0T1]=[M0L1T0]a[M1L1T0]b[M1L1T2]c\Rightarrow [{M^0}{L^0}{T^{ - 1}}] = {[{M^0}{L^1}{T^0}]^a}{[{M^1}{L^{ - 1}}{T^0}]^b}{[{M^1}{L^1}{T^{ - 2}}]^c}

Comparing the LHS and RHS for MM , we get:

0=a×0+b×1+c×1\Rightarrow 0 = a \times 0 + b \times 1 + c \times 1

0=b+c\Rightarrow 0 = b + c [Eq. 1]

Now doing the same for L, gives us:

0=a×1+b×1+c×1\Rightarrow 0 = a \times 1 + b \times - 1 + c \times 1

0=ab+c\Rightarrow 0 = a - b + c [Eq. 2]

Again, for TT , we get:

1=a×0+b×0+c×2\Rightarrow - 1 = a \times 0 + b \times 0 + c \times - 2

1=2c\Rightarrow - 1 = - 2c

This gives us c=12c = \dfrac{1}{2} .

Putting this value of cc in Eq. 1, we have:

0=b+12\Rightarrow 0 = b + \dfrac{1}{2}

b=12\Rightarrow b = - \dfrac{1}{2}

Again, substituting both bb and cc in Eq. 2:

0=a(12)+12\Rightarrow 0 = a - \left( { - \dfrac{1}{2}} \right) + \dfrac{1}{2}

0=a+12+12\Rightarrow 0 = a + \dfrac{1}{2} + \dfrac{1}{2}

This gives us the value of a=1a = - 1 .

Combining the values of aa , bb and cc and putting in our original equation gives us:

v=kl1m1/2T1/2\Rightarrow v = k{l^{ - 1}}{m^{ - 1/2}}{T^{1/2}}

v=klTm\Rightarrow v = \dfrac{k}{l}\sqrt {\dfrac{T}{m}}

Hence, the formula for frequency of the string is v=klTmv = \dfrac{k}{l}\sqrt {\dfrac{T}{m}} where kk is a constant.

Note:

The constant of proportionality is determined by experimental results in most cases. It gives us an idea about the extent to which a quantity increases or decreases if we keep other variables with unit value. It is also known as the ratio between two proportional quantities.