JEE MAIN - Physics (2010 - No. 20)
The tension in the string is
$$4.0N$$
$$12.5$$ $$N$$
$$0.5$$ $$N$$
$$6.25$$ $$N$$
Explanation
$$y = 0.02\left( m \right)\sin \left[ {2\pi \left( {{t \over {0.04\left( s \right)}}} \right) - {x \over {0.50\left( m \right)}}} \right]$$
But $$y = a\sin \left( {\omega t - kx} \right)$$
$$\therefore$$ $$\omega = {{2\pi } \over {0.04}} \Rightarrow v = {1 \over {0.04}} = 25\,Hz$$
$$k = {{2\pi } \over {0.50}} \Rightarrow \lambda = 0.5m$$
$$\therefore$$ velocity, $$v = v\lambda = 25 \times 0.5\,m/s = 12.5\,m/s$$
Velocity on a string is given by
$$v = \sqrt {{T \over \mu }} $$
$$\therefore$$ $$T = {v^2} \times \mu = {\left( {12.5} \right)^2} \times 0.04 = 6.25\,N$$
But $$y = a\sin \left( {\omega t - kx} \right)$$
$$\therefore$$ $$\omega = {{2\pi } \over {0.04}} \Rightarrow v = {1 \over {0.04}} = 25\,Hz$$
$$k = {{2\pi } \over {0.50}} \Rightarrow \lambda = 0.5m$$
$$\therefore$$ velocity, $$v = v\lambda = 25 \times 0.5\,m/s = 12.5\,m/s$$
Velocity on a string is given by
$$v = \sqrt {{T \over \mu }} $$
$$\therefore$$ $$T = {v^2} \times \mu = {\left( {12.5} \right)^2} \times 0.04 = 6.25\,N$$
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