The fundamental frequency f of a stretched vibrating string depends on the tension force F… — JAMB UTME

The fundamental frequency $f$ of a stretched vibrating string depends on the tension force $F$ , the length of the string $l$ , and its linear mass density $\mu$ (mass per unit length) according to the relation $f = k F^a l^b \mu^c$ , where $k$ is a dimensionless constant. Which of the following sets of exponents $(a, b, c)$ correctly satisfies dimensional homogeneity?

$(\frac{1}{2}, -1, -\frac{1}{2})$ Answer
B
$(1, -1, -1)$
C
$(\frac{1}{2}, 1, \frac{1}{2})$
D
$(-\frac{1}{2}, -1, \frac{1}{2})$

Answer

The correct set of exponents is

(\frac{1}{2}, -1, -\frac{1}{2})

Equating the base dimensions on both sides gives

T^{-1} = M^{a+c} L^{a+b-c} T^{-2a}

. Solving for the powers gives

a = \frac{1}{2}

b = -1

, and

c = -\frac{1}{2}

, matching the option specifying

(\frac{1}{2}, -1, -\frac{1}{2})

Step-by-Step Solution

Express the dimensions of all physical quantities involved in fundamental dimensions

M

L

, and

T

[f] = T^{-1}

[F] = M L T^{-2}

[l] = L

, and

[\mu] = M L^{-1}

Linear mass density

\mu

is mass per unit length (

\text{mass}/\text{length}

Substitute dimensions into the given formula

f = k F^a l^b \mu^c

(ignoring the dimensionless constant

k

T^{-1} = (M L T^{-2})^a (L)^b (M L^{-1})^c = M^{a+c} L^{a+b-c} T^{-2a}

Dimensional homogeneity requires both sides of the equation to have identical dimensional powers.

Equate exponents of

M

L

, and

T

from both sides to form algebraic equations.

For

T

-2a = -1 \implies a = \frac{1}{2}

. For

M

a + c = 0 \implies c = -a = -\frac{1}{2}

. For

L

a + b - c = 0 \implies \frac{1}{2} + b - (-\frac{1}{2}) = 0 \implies b + 1 = 0 \implies b = -1

Solving the system yields the unique set of exponents

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

Key Concept

Dimensional Analysis and Method of Dimensions

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic