Question 209: Waves

Question

The equation of a wave on a string of linear mass density $$0.04\,\,kg\,{m^{ - 1}}$$ is given by 
$$$y = 0.02\left( m \right)\,\sin \left[ {2\pi \left( {{t \over {0.04\left( s \right)}} - {x \over {0.50\left( m \right)}}} \right)} \right].$$$
The tension in the string is

Accepted Answer

$$y = 0.02\left( m ight)\sin \left[ {2\pi \left( {{t \over {0.04\left( s ight)}}} ight) - {x \over {0.50\left( m ight)}}} ight]$$
But $$y = a\sin \left( {\omega t - kx} ight)$$
$$	herefore$$ $$\omega  = {{2\pi } \over {0.04}} \Rightarrow v = {1 \over {0.04}} = 25\,Hz$$
$$k = {{2\pi } \over {0.50}} \Rightarrow \lambda  = 0.5m$$
$$	herefore$$ velocity, $$v = v\lambda  = 25 	imes 0.5\,m/s = 12.5\,m/s$$
Velocity on a string is given by
$$v = \sqrt {{T \over \mu }} $$
$$	herefore$$ $$T = {v^2} 	imes \mu  = {\left( {12.5} ight)^2} 	imes 0.04 = 6.25\,N$$

Answer

$$4.0N$$

Answer

$$12.5$$ $$N$$

Answer

$$0.5$$ $$N$$

Answer

$$6.25$$ $$N$$

Waves

The equation of a wave on a string of linear mass density $0.04\,\,kg\,{m^{ - 1}} is given by y = 0.02\left( m \right)\,\sin \left[ {2\pi \left( {{t \over {0.04\left( s \right)}} - {x \over {0.50\left( m \right)}}} \right)} \right].$$ The tension in the string is

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