Login/Sign Up
Home » A string clamped at both its ends is stretched out, it is then made to vibrate in its fundamental mode at a frequency of . The linear mass density of the string is m and its mass is kg. Calculate:(i) the velocity of a transverse wave on the string,(ii) the tension in the string.
A string clamped at both its ends is stretched out, it is then made to vibrate in its fundamental mode at a frequency of 45 \mathrm{~Hz}. The linear mass density of the string is 4.0 \times 10^{-2} \mathrm{~kg} / m and its mass is 2 \times 10^{-2} kg. Calculate:
(i) the velocity of a transverse wave on the string,
(ii) the tension in the string.
CBSE | Class 11 | NCERT | Physics | Waves
A string clamped at both its ends is stretched out, it is then made to vibrate in its fundamental mode at a frequency of 45 \mathrm{~Hz}. The linear mass density of the string is 4.0 \times 10^{-2} \mathrm{~kg} / m and its mass is 2 \times 10^{-2} kg. Calculate:
(i) the velocity of a transverse wave on the string,
(ii) the tension in the string.

Mass of the string is given as m=2 \times 10^{-2} \mathrm{~kg}

Linear density of the string is given s =4 \times 10^{-2} \mathrm{~kg}

Frequency is given as \mathrm{v}_{\mathrm{F}}=45 \mathrm{~Hz}

length of the wire can be calculated as

\mathrm{m} / \mathrm{\mu} =\left(2 \times 10^{-2}\right) /\left(4 \times 10^{-2}\right)=0.5 \mathrm{~m}

As, \lambda=2 \mathrm{l} / \mathrm{n}

Where, n is the number of nodes in the wire.

For fundamental node, n=1, we have

\begin{array}{l} =>\lambda=2 \mid \\ =2 \times 0.5=1 \mathrm{~m} \end{array}

(i) As a result, speed of the transverse wave will be,

v=\lambda v_{F} =1 \times 45=45 \mathrm{~m} / \mathrm{s}

(ii) Tension in the string will be,

\mu \mathrm{v}^{2} =4 \times 10^{-2} \times 45=81 \mathrm{~N}

i

More Material