    Question

# A 2⋅00 m-long rope, having a mass of 80 g, is fixed at one end and is tied to a light string at the other end. The tension in the string is 256 N. (a) Find the frequencies of the fundamental and the first two overtones. (b) Find the wavelength in the fundamental and the first two overtones.

Open in App
Solution

## Given: Length of the long rope (L) = 2.00 m Mass of the rope = 80 g = 0.08 kg Tension (T) = 256 N Linear mass density, m $=\frac{0.08}{2}=0.04\mathrm{kg}/\mathrm{m}$ $\mathrm{Tension},T=256\mathrm{N}\phantom{\rule{0ex}{0ex}}\mathrm{Wave}\mathrm{velocity},v=\sqrt{\frac{\mathrm{T}}{m}}\phantom{\rule{0ex}{0ex}}⇒v=\sqrt{\left(\frac{256}{0.04}\right)}=\frac{160}{2}\phantom{\rule{0ex}{0ex}}⇒v=80\mathrm{m}/\mathrm{s}$ For fundamental frequency: $L=\frac{\lambda }{4}\phantom{\rule{0ex}{0ex}}⇒\lambda =4L=4×2=8\mathrm{m}\phantom{\rule{0ex}{0ex}}⇒f=\frac{v}{\lambda }=\frac{80}{8}=10\mathrm{Hz}$ (a) The frequency overtones are given below: $1\mathrm{st}\mathrm{overtone}=3f=30\mathrm{Hz}\phantom{\rule{0ex}{0ex}}2\mathrm{nd}\mathrm{overtone}=5f=50\mathrm{Hz}$ (b) $\lambda =4l=4×2=8\mathrm{m}$ $\therefore {\lambda }_{1}=\frac{v}{{f}_{1}}=\frac{80}{30}=2.67\mathrm{m}\phantom{\rule{0ex}{0ex}}{\lambda }_{2}=\frac{v}{{f}_{2}}=\frac{80}{50}=1.6\mathrm{m}$ Hence, the wavelengths are 8 m, 2.67 m and 1.6 m, respectively.  Suggest Corrections  0      Similar questions  Related Videos   Technologies
PHYSICS
Watch in App  Explore more