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 \( b = \frac{1}{2} \).

Answer 3

The equation \( v = (\text{constant}) F^a \left( \dfrac{m}{L} \right)^b \) must balance dimensionally on both sides. Speed has dimensions \( [v] = [L][T]^{-1} \), tension has dimensions \( [F] = [M][L][T]^{-2} \), and mass per unit length has dimensions \( \left[ \dfrac{m}{L} \right] = [M][L]^{-1} \). Let's check each proposed pair of exponents against this requirement.

1. \( a = \frac{1}{2}, b = \frac{1}{2} \): This is the pairing built from a square-root dependence on tension together with a square-root dependence on mass per length, the same structure the wave speed on a string actually follows.
2. \( a = 2, b = -1 \): Here \( a + b = 1 \), so the right side carries a leftover mass dimension \( [M]^1 \) that cannot match the massless left side. This pairing overstates how strongly the speed grows with tension.
3. \( a = \frac{1}{2}, b = -1 \): Here \( a + b = -\frac{1}{2} \), leaving a fractional negative mass dimension on the right side with no counterpart on the left. This pairing understates the tension dependence relative to what dimensional balance requires.
4. \( a = 1, b = \frac{1}{2} \): Here \( a + b = \frac{3}{2} \), even further from the mass-free requirement, so this pairing is inconsistent too.

For the mass dimension to cancel we need \( a + b = 0 \). For the length dimension to give exactly one power of \( L \) we need \( a - b = 1 \), and for the time dimension we need \( -2a = -1 \), so \( a = \frac{1}{2} \) and, from \( a + b = 0 \), \( b = -\frac{1}{2} \). Among the four choices, only the first pairs a square-root rise with tension against a square-root fall with mass per length, matching this relationship.

So the correct answer is \( a = \frac{1}{2}, b = -\frac{1}{2} \), corresponding to the first option.