Test Index

Waves

© examsnet.com
Question : 14 of 27
Marks: +1, -0
A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5×10−2kg3.5 \times 10^{-2} \mathrm{kg} and its linear mass density is 4.0×10−2kgm−14.0 \times 10^{-2} \mathrm{kg} \mathrm{m}^{-1}. What is (a) the speed of a transverse wave on the string, and (b) the tension in the string?
Solution:  
Here, υ=45\upsilon = 45 Hz, M=3.5×10−2M = 3.5 \times 10^{-2} kg ;
mass/length =μ=4.0×10−2kgm−1\mu = 4.0 \times 10^{-2} \mathrm{kg} \mathrm{m}^{-1}
∴l=Mμ=3.5×10−24.0×10−2=78m\therefore l = \frac{M}{\mu} = \frac{3.5 \times 10^{-2}}{4.0 \times 10^{-2}} = \frac{7}{8} \mathrm{m}
As wire vibrates in its fundamental mode
λ2=l=78\frac{\lambda}{2} = l = \frac{7}{8}
∴λ=74m=1.75m\therefore \lambda = \frac{7}{4} \mathrm{m} = 1.75 \mathrm{m}
The speed of the transverse wave v=υλv = \upsilon \lambda =45×1.75=78.75ms−1=45 \times 1.75 = 78.75 \mathrm{m} \mathrm{s}^{-1}
(b) As, v=Tμv = \sqrt{ \frac{T}{\mu} }
T=v2×μT = v^{2} \times \mu =(78.75)2×4.0×10−2= (78.75)^{2} \times 4.0 \times 10^{-2} =248.06N= 248.06 \mathrm{N}
© examsnet.com
Go to Question:
Prev Question
Next Question