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Q14.14(b):
A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 (\times) 10^{-2} kg and its linear mass density is 4.0 (\times) 10^{-2} kg (m^{-1}). What is the tension in the string?

Solution :

Step 1: Given Variables & Initial Setup

To determine the tension in the wire, we define the known physical parameters provided in the problem statement:

  • Fundamental Frequency ($f$): $45\text{ Hz}$
  • Total Mass of the wire ($M$): $3.5 \times 10^{-2}\text{ kg}$
  • Linear Mass Density ($\mu$): $4.0 \times 10^{-2}\text{ kg m}^{-1}$

Step 2: Determining the Length of the Wire

The linear mass density $\mu$ is defined as the total mass $M$ divided by the length $L$. Therefore, we calculate $L$ as follows:

$\mu = \frac{M}{L} \implies L = \frac{M}{\mu}$

$L = \frac{3.5 \times 10^{-2}\text{ kg}}{4.0 \times 10^{-2}\text{ kg m}^{-1}} = 0.875\text{ m}$

Step 3: Relationship Between Frequency, Tension, and Length

For a string fixed at both ends vibrating in its fundamental mode (first harmonic), the length $L$ corresponds to half a wavelength ($\lambda/2$). [Per the wave equation for a stretched string]:

$\lambda = 2L$

The velocity of a wave on a stretched string is given by $v = \sqrt{\frac{T}{\mu}}$, where $T$ is the tension. Since $v = f\lambda$, we substitute for $\lambda$ and $v$:

$f(2L) = \sqrt{\frac{T}{\mu}}$

Step 4: Algebraic Derivation for Tension ($T$)

Squaring both sides of the equation to isolate $T$:

$4f^2L^2 = \frac{T}{\mu}$

$T = 4f^2L^2\mu$

Alternatively, substituting $L = M/\mu$ into the expression:

$T = 4f^2 \left(\frac{M}{\mu}\right)^2 \mu = \frac{4f^2M^2}{\mu}$

Step 5: Numerical Calculation

Substituting the values into the derived formula:

$T = 4 \times (45\text{ Hz})^2 \times (0.875\text{ m})^2 \times (4.0 \times 10^{-2}\text{ kg m}^{-1})$

$T = 4 \times 2025 \times 0.765625 \times 0.04$

$T = 8100 \times 0.030625$

$T = 248.0625\text{ N}$

Fundamental Mode ($\lambda = 2L$) Support 1 Support 2

Final Solution: The tension in the wire is approximately $248.06\text{ N}$.


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