1.

A stationary wave is described by the equation `y=10sin(x//3)cos(15pit),` where x is in cm and time t is in second. Calculate frequency and velocity of component waves. Also, find the amplitude of stationary waves at antinodes and find the position of first antinode.

Answer» The given eqn. is
`y=10sin(x//3)cos15pit`
Comparing with the standard form of eqn.
`y=2sinkxcosomegat, ` we get
`2r=10cm, k=(2ppi)/(lambda)=(1)/(3), omega=15pi`
`lambda=6pi=6xx3.14=18.84cm`
`omega =2piv=15pi,v=7.5Hz`
`upsilon=vlambda=7.5xx18.84=141.3cms^(-1)`
Amplitude of stationary wave `=10sin((x)/(3))`
At antinode, amplitude is maximum, `sin((x)/(3))=max. :. (x)/(3)=pi//2 x=(3pi)/(2)=4.71cm`
Amplitude of stationary wave at antinode
`=10sin(x//3)=10xx1cm=0.1m`


Discussion

No Comment Found

Related InterviewSolutions