A string of length 1 m and mass $$2 \times 10^{-5}$$ 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.

For a string fixed at both ends, the frequency of the $$n^{\text{th}}$$ harmonic is given by
$$f_n = \frac{n\,v}{2L}$$
where $$v$$ is the wave speed on the string and $$L$$ is its length.

Successive harmonics correspond to $$n$$ and $$n+1$$, so the difference between their frequencies is
$$\Delta f = f_{n+1} - f_n = \frac{(n+1)v}{2L} - \frac{nv}{2L} = \frac{v}{2L}$$ $$-(1)$$

The two given successive frequencies are 750 Hz and 1000 Hz, hence
$$\Delta f = 1000 - 750 = 250\ \text{Hz}$$

Substituting $$L = 1\ \text{m}$$ into $$(1)$$:
$$v = 2L\,\Delta f = 2 \times 1 \times 250 = 500\ \text{m s}^{-1}$$

The wave speed on a stretched string is also related to the tension $$T$$ and the linear mass density $$\mu$$ by
$$v = \sqrt{\frac{T}{\mu}}$$ $$\Longrightarrow$$ $$T = \mu v^{2}$$ $$-(2)$$

The string’s mass is $$2 \times 10^{-5}\ \text{kg}$$ and its length is $$1\ \text{m}$$, so
$$\mu = \frac{\text{mass}}{\text{length}} = \frac{2 \times 10^{-5}}{1} = 2 \times 10^{-5}\ \text{kg m}^{-1}$$

Substituting $$\mu$$ and $$v = 500\ \text{m s}^{-1}$$ into equation $$(2)$$:
$$T = (2 \times 10^{-5}) \times (500)^{2}$$
$$T = 2 \times 10^{-5} \times 250000$$
$$T = 5\ \text{N}$$

Therefore, the tension in the string is 5 N.