Question 286: Waves

Question

A string of mass $0.1 \mathrm{kgm}^{-1}$ has length 0.9 m . It is fixed at both ends and stretched such that it has a tension of 40 N . The string vibrates in three segments with amplitude 0.3 cm . The amplitude (maximum) of the particle velocity is (in $\mathrm{m} / \mathrm{s}$)

Accepted Answer

Given values:
Mass per unit length, $\mu = 0.1\,\mathrm{kg}/\mathrm{m}$
Length of string, $L = 0.9\,\mathrm{m}$
Tension, $T = 40\,\mathrm{N}$
Number of segments (loops/antinodes), $n = 3$
Amplitude, $A = 0.3\,\mathrm{cm} = 0.003\,\mathrm{m}$
Let's find the
maximum particle velocity amplitude
.
Step 1: Find the frequency of vibration
For a string fixed at both ends:
$ 	ext{Wavelength for the nth harmonic: } \lambda_n = \frac{2L}{n} $
$ \lambda_3 = \frac{2 	imes 0.9}{3} = \frac{1.8}{3} = 0.6\,\mathrm{m} $
Wave speed, $v$:
$ v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{40}{0.1}} = \sqrt{400} = 20\,\mathrm{m/s} $
Frequency, $f$:
$ f = \frac{v}{\lambda_3} = \frac{20}{0.6} = \frac{200}{6} = 33.33\,\mathrm{Hz} $
Angular frequency, $\omega$:
$ \omega = 2\pi f = 2\pi 	imes 33.33 = 66.66\pi \,\mathrm{rad/s} $
Step 2: Relation between particle velocity and amplitude
The
maximum particle velocity
at an antinode is:
$ v_	ext{max} = \omega A $
Plug in the values:
$A = 0.003\,\mathrm{m}$
$\omega = 2\pi f = 2\pi 	imes 33.33 = 66.66\pi$
$ v_{	ext{max}} = \omega A = 66.66\pi 	imes 0.003 $
$ = (66.66 	imes 0.003) \pi $
$ = 0.2 \pi \;\mathrm{m/s} $
Step 3: Match with the given options
Let us compare $0.2\pi$ with the options:
$ \frac{\pi}{5} = 0.2\pi $
Final Answer
$ \boxed{\frac{\pi}{5}\ \mathrm{m/s}} $
So, the correct option is
C
.

Answer

$\frac{\pi}{2}$

Answer

$\frac{\pi}{3}$

Answer

$\frac{\pi}{5}$

Answer

$\frac{\pi}{6}$

Waves

A string of mass 0.1 \mathrm{kgm}^{-1} has length 0.9 m . It is fixed at both ends and stretched such that it has a tension of 40 N . The string vibrates in three segments with amplitude 0.3 cm . The amplitude (maximum) of the particle velocity is (in \mathrm{m} / \mathrm{s})

Choose the correct answer: