Question

The speed \( v \) of a wave on a string depends on the tension \( F \) in the string and the mass per unit length \( m/L \) of the string. If it is known that [F] = [ML][T]–2, the values of the constants \( a \) and \( b \) in the following equation for the speed of a wave on a string are: \[ v = (\text{constant}) F^a \left( \frac{m}{L} \right)^b \]

• A. \( a = \frac{1}{2}, b = \frac{1}{2} \)
• B. \( a = 2, b = -1 \)
• C. \( a = \frac{1}{2}, b = -1 \)
• D. \( a = 1, b = \frac{1}{2} \)

Hint:

The speed of a wave on a string depends on the tension and mass per unit length, with the speed being proportional to the square root of the tension and inversely proportional to the square root of the mass per unit length.

Answer 1

The Correct Option is A

The speed \( v \) of a wave on a string is given by the equation \( v = (\text{constant}) F^a \left( \frac{m}{L} \right)^b \). We need to find the values of constants \( a \) and \( b \) such that this equation is dimensionally consistent. 

We begin by analyzing the dimensions involved. The dimension of speed \( v \) is \([v] = [LT^{-1}]\). Given, the dimension of tension \( F \) is \([F] = [ML][T]^{-2} = [M][L][T]^{-2}\). The mass per unit length \(\frac{m}{L}\) has the dimension \([\frac{m}{L}] = [M][L]^{-1}\).

Substitute these into the equation:

\([v] = [F]^a \left[\frac{m}{L}\right]^b = ([M][L][T]^{-2})^a ([M][L]^{-1})^b\)

Solving for the dimensions, we get:

\([L][T]^{-1} = [M]^a[L]^a[T]^{-2a}[M]^b[L]^{-b}\)

Combining the dimensions, we have:

\([L][T]^{-1} = [M]^{a+b}[L]^{a-b}[T]^{-2a}\)

Set the powers of \(M\), \(L\), and \(T\) equal to each other:

• For \(M\): \(a + b = 0\)
• For \(L\): \(a - b = 1\)
• For \(T\): \(-2a = -1\)

From \(-2a = -1\), we get \(a = \frac{1}{2}\).

Substitute \(a = \frac{1}{2}\) into \(a + b = 0\):

\(\frac{1}{2} + b = 0 \Rightarrow b = -\frac{1}{2}\).

Checking with other relation \(a - b = 1\):

\(\frac{1}{2} - (-\frac{1}{2}) = 1\), which is correct.

Thus, the values of \(a\) and \(b\) are \(\frac{1}{2}\) and \(-\frac{1}{2}\), respectively. Hence, the correct option is:

\(a = \frac{1}{2}, b = -\frac{1}{2}\).

Answer 2

The speed of the wave on the string is given by: \[ v = \sqrt{\frac{F}{\mu}} \] Where \( \mu = \frac{m}{L} \) is the mass per unit length of the string. Therefore, the speed is proportional to the square root of the tension \( F \) and the inverse square root of the mass per unit length. Thus, \( a = \frac{1}{2} \) and \( b = \frac{1}{2} \).