Question

A string of mass 150 grams fixed of both ends of length 1.5 m is vibrating in 2nd overtone. Maximum poss speed of any particle on string is 0.8 $\pi$ m/s. Particles are oscillating with frequency 20 Hz

• A. Tension in the string is 40 N
• B. Maximum speed of a particle which is $\frac{25}{3}$ cm from one end is 40 $\pi$ cm/s
• C. Maximum speed of a particle which is $\frac{25}{3}$ cm from one end is 20 $\pi$ cm/s
• D. All the particles of string are in same phase

Answer 1

Answer: (A), (B)

Options A and B are correct.

Answer 2

1. Tension Calculation:

   • Total mass $m = 0.150$ kg and length $L = 1.5$ m so that mass per unit length is $$\mu = \frac{m}{L} = \frac{0.150}{1.5} = 0.1\, \text{kg/m}.$$
   • The string is vibrating in the 2nd overtone, which corresponds to the 3rd harmonic. For fixed ends, the wavelength in the $n$th harmonic is $$\lambda = \frac{2L}{n},\quad \text{with } n=3 \quad \Rightarrow \quad \lambda = \frac{2 \times 1.5}{3} = 1 \text{ m}.$$
   • Given the frequency $f = 20$ Hz, the wave speed is $$v = f\, \lambda = 20 \times 1 = 20\, \text{m/s}.$$
   • Then the tension is $$T = \mu\, v^2 = 0.1 \times 20^2 = 40\, \text{N}.$$

   This confirms option A.

2. Amplitude and Maximum Speed of a Particle:

   • The maximum speed of a particle at an antinode is given by $$v_{\text{max}} = \omega A,$$ where the angular frequency is $$\omega = 2\pi f = 40\pi\, \text{rad/s}.$$
   • We are told that the maximum possible speed on the string is $0.8\pi\, \text{m/s}$ (attained at an antinode). Thus, $$40\pi \,A_{\text{max}} = 0.8\pi \quad \Rightarrow \quad A_{\text{max}} = \frac{0.8\pi}{40\pi} = 0.02\, \text{m}.$$
   • In a standing wave (3rd harmonic) on a string fixed at both ends, the displacement can be written as $$y(x,t) = A_{\text{max}} \sin\left(\frac{3\pi x}{L}\right)\cos(\omega t).$$ Notice that for $L = 1.5\, \text{m}$, $$\frac{3\pi x}{1.5} = 2\pi x.$$ So, $$y(x,t) = A_{\text{max}} \sin (2\pi x)\cos(40\pi t).$$
   • The local amplitude at a distance $x$ is $A(x) = A_{\text{max}} \sin(2\pi x)$. For $x = \frac{25}{3}$ cm $= \frac{25}{300} = \frac{1}{12}\, \text{m}$, $$A\left(\frac{1}{12}\right) = 0.02 \times \sin\left(2\pi \cdot \frac{1}{12}\right) = 0.02 \times \sin\left(\frac{\pi}{6}\right) = 0.02 \times \frac{1}{2} = 0.01\, \text{m}.$$
   • The maximum speed at that point is $$v_{\text{max}}(x)= \omega\, A(x) = 40\pi \times 0.01 = 0.4\pi\, \text{m/s}.$$ Converting $0.4\pi\, \text{m/s}$ to cm/s, $$0.4\pi \times 100 = 40\pi\, \text{cm/s}.$$

   This confirms option B. Option C stating $20\pi$ cm/s is incorrect.

3. Phase Relation:

   • In a standing wave, different points on the string do not oscillate in the same phase (only points at the same antinode might be in phase, but the phase varies with position). Thus, option D is false.