Question

a tring of length 1.5m is oscillatin is 2nd oevrtone. the tension is 100N and mass is 15gm amplitude is 1mm at antinode

Answer 1

• Wave speed: 100 m/s
• Frequency: 100 Hz
• Maximum particle speed at the antinode: 0.2π m/s

Answer 2

We are given a string with

• Length, $L = 1.5$ m
• Mass, $m = 15$ g = 0.015 kg ⟹ mass per unit length, $\mu = m/L = 0.015/1.5 = 0.01$ kg/m
• Tension, $T = 100$ N
• The string oscillates in the 2nd overtone (i.e. the 3rd harmonic)
• Amplitude at the antinode, $A = 1$ mm = 0.001 m

Step 1. Determine the wavelength and wave speed

For a string fixed at both ends vibrating in the $n^{th}$ harmonic, the wavelength is
$\lambda = 2L/n$
Since the 2nd overtone corresponds to $n = 3$,
$\lambda = (2 \times 1.5) / 3 = 1$ m

The speed of the wave is given by
$v = \sqrt{T/\mu} = \sqrt{100/0.01} = \sqrt{10000} = 100$ m/s

Step 2. Find the frequency

The relation between wave speed, frequency, and wavelength is
$v = f\lambda$ ⟹ $f = v/\lambda = 100/1 = 100$ Hz

Step 3. Find the maximum speed of a particle at the antinode

In a standing wave the maximum transverse speed of a particle is
$v_{max} = \omega A$, where $\omega = 2\pi f$
Thus,
$\omega = 2\pi \times 100 = 200\pi$ rad/s
$v_{max} = (200\pi)(0.001) = 0.2\pi$ m/s