The speed v of a transverse wave traveling along a stretched string under tension T with m… — JAMB UTME

The speed $v$ of a transverse wave traveling along a stretched string under tension $T$ with mass per unit length $\mu$ is given by $v = k T^x \mu^y$ , where $k$ is a dimensionless constant. What is the numerical value of the exponent $x$ ?

Answer: 0.5

Answer

The numerical value of the exponent

x

is 0.5.

Applying the principle of dimensional homogeneity, the dimensions on both sides must match. Speed

[v] = L T^{-1}

, tension force

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

, and mass per unit length

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

. Substituting these into

v = k T^x \mu^y

yields

M^0 L^1 T^{-1} = M^{x+y} L^{x-y} T^{-2x}

. Comparing the exponents of

T

gives

-2x = -1

, leading to

x = 0.5

Step-by-Step Solution

Determine the dimensions of speed

v

, tension force

T

, and linear density

\mu

[v] = L T^{-1}

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

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

Tension is a force (

F=ma

) with dimensions

[M L T^{-2}]

, and

\mu

is mass per unit length (

m/l

) with dimensions

[M L^{-1}]

Substitute the dimensional formulas into the equation

v = k T^x \mu^y

and collect powers of base dimensions.

M^0 L^1 T^{-1} = M^{x+y} L^{x-y} T^{-2x}

Combining exponents for base dimensions

M

L

, and

T

allows applying the principle of dimensional homogeneity.

Equate the exponent of

T

on both sides to solve for

x

-2x = -1 \implies x = 0.5

The exponent of

T

on the left side is

-1

and on the right side is

-2x

Key Concept

Dimensional Analysis and Determination of Exponents

Estimated Time:1m 30s

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