Question

A stretched string of mass 20 g vibrates with a frequency of 30 Hz in its fundamental
mode and the supports are 40 cm apart. The amplitude of vibrations at the antinode is
4 cm. Calculate the velocity of propagation of the wave in the string as well as the
tension in it.

Answer 1

Answer on Question 66261, Physics, Mechanics, Relativity

Question:

A stretched string of mass $20\,g$ vibrates with a frequency of $30\,Hz$ in its fundamental mode and the supports are $40\,cm$ apart. The amplitude of vibrations at the antinode is $4\,cm$. Calculate the velocity of propagation of the wave in the string as well as the tension in it.

Solution:

a) We can find the velocity of propagation of the wave in the string from the wave speed formula:

$$v = f \lambda,$$

here, $v$ is the velocity of propagation of the wave in the string, $f$ is the frequency, $\lambda$ is the wavelength.

If the length of the string is $L$, the fundamental mode is the one produced by the vibration whose nodes are the two ends of the string, so $L$ is half of the wavelength of the fundamental mode. Then, the wavelength of the fundamental mode will be equal to $\lambda = 2L$ and we can calculate the velocity of propagation of the wave in the string:

$$v = f \lambda = f 2 L = 3 0\,Hz \cdot 2 \cdot 0.4\,m = 24\,\frac{m}{s}.$$

b) We can find the tension in the string from the formula:

$$v = \sqrt{\frac{T}{\mu}},$$

here, $v$ is the velocity of propagation of the wave in the string, $T$ is the tension in the string, $\mu = M/L$ is the mass per unit length of the string.

Then, from this formula we can calculate the tension in the string:

$$v^2 = \frac{T}{\mu},$$ $$T = v^2 \mu = v^2 \frac{M}{L} = \left(24\,\frac{m}{s}\right)^2 \cdot \frac{0.02\,kg}{0.4\,m} = 28.8\,N.$$

Answer:

a) $v = 24\ \frac{m}{s}$.

b) $T = 28.8\ N$.

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