A string of length 1 m and mass 2 × 10−5 kg is under tension 𝑇. When the string vibrates, two successive harmonics are found to occur at frequencies 750 Hz and 1000 Hz. The value of tension 𝑇 is _____ Newton.

 To determine the tension 

$T$ in the string, we'll follow these steps:

Given:

• Length of the string, $L=1\text{m}$

• Mass of the string, $m=2\times 10^{-5}\text{kg}$

• Two successive harmonic frequencies: $f_n=750\text{Hz}$ and $f_{n+1}=1000\text{Hz}$

Step 1: Understand the relationship between harmonics and frequency

For a string fixed at both ends, the frequency of the $n$-th harmonic is given by:

$$f_n=\frac{n}{2L}\sqrt{\frac{T}{\mu }}$$

where:

• $n$ is the harmonic number (1, 2, 3, ...)

• $\mu$ is the linear mass density of the string ($\mu =\frac{m}{L}$)

• $T$ is the tension in the string

Step 2: Express the frequencies of two successive harmonics

Let the two successive harmonics be $n$ and $n+1$. Their frequencies are:

$$f_n=\frac{n}{2L}\sqrt{\frac{T}{\mu }}$$$$f_{n+1}=\frac{n+1}{2L}\sqrt{\frac{T}{\mu }}$$

Step 3: Take the ratio of the two frequencies

$$\frac{f_{n+1}}{f_n}=\frac{\frac{n+1}{2L}\sqrt{\frac{T}{\mu }}}{\frac{n}{2L}\sqrt{\frac{T}{\mu }}}=\frac{n+1}{n}$$

Substitute the given frequencies:

$$\frac{1000}{750}=\frac{n+1}{n}$$

Simplify:

$$\frac{4}{3}=\frac{n+1}{n}$$

Solve for $n$:

$$4n=3(n+1)$$$$4n=3n+3$$$$n=3$$

Step 4: Use the frequency formula to solve for $T$

Now that we know $n=3$, substitute $f_n=750\text{Hz}$ into the frequency formula:

$$750=\frac{3}{2\cdot 1}\sqrt{\frac{T}{\mu }}$$

Simplify:

$$750=\frac{3}{2}\sqrt{\frac{T}{\mu }}$$$$\sqrt{\frac{T}{\mu }}=\frac{750\cdot 2}{3}=500$$

Square both sides:

$$\frac{T}{\mu }=500^2=250000$$

Now, substitute $\mu =\frac{m}{L}=\frac{2\times 10^{-5}}{1}=2\times 10^{-5}\text{kg/m}$:

$$T=250000\cdot \mu =250000\cdot 2\times 10^{-5}$$$$T=5\text{N}$$

Final Answer:

The tension $T$ in the string is:

$$\boxed{{\displaystyle 5}}$$