Calculate Tension in Vibrating String Given Successive Harmonics

To find the tension $T$ in the vibrating string, we analyze the condition for standing waves on a string fixed at both ends.

1. Linear Mass Density ($\mu$): The linear mass density of the string is given by: $\mu = \frac{\text{Mass}}{\text{Length}} = \frac{2 \times 10^{-5}\text{ kg}}{1\text{ m}} = 2 \times 10^{-5}\text{ kg/m}$

2. Fundamental Frequency ($f_1$): For a string fixed at both ends, the frequencies of successive harmonics are consecutive integer multiples of the fundamental frequency $f_1$. Let the two successive frequencies be $f_n$ and $f_{n+1}$: $f_n = n f_1 = 750\text{ Hz}$ $f_{n+1} = (n+1) f_1 = 1000\text{ Hz}$

Subtracting the two equations gives the fundamental frequency: $f_1 = f_{n+1} - f_n = 1000\text{ Hz} - 750\text{ Hz} = 250\text{ Hz}$

(Note: We can confirm that $n = \frac{750}{250} = 3$, so these correspond to the $3^{\text{rd}}$ and $4^{\text{th}}$ harmonics).

3. Wave Speed ($v$): The relationship between the fundamental frequency, the length of the string $L$, and the wave speed $v$ is: $f_1 = \frac{v}{2L}$

Rearranging to solve for wave speed $v$: $v = 2 L f_1 = 2 \times 1\text{ m} \times 250\text{ Hz} = 500\text{ m/s}$

4. Tension ($T$): The speed of a transverse wave on a stretched string is related to the tension $T$ and linear mass density $\mu$ by: $v = \sqrt{\frac{T}{\mu}}$

Squaring both sides and solving for $T$: $T = \mu v^2$

Substituting the known values: $T = (2 \times 10^{-5}\text{ kg/m}) \times (500\text{ m/s})^2$ $T = 2 \times 10^{-5} \times 250000$ $T = 5\text{ N}$