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A displacement wave is represented by `xi=0.25xx10^(-3)sin(500t=0.025x)` Deduce (i) amplitude (ii) period (iii) angular frequency (iv)wavelength (v) amplitude of particle velocity (vi) amplitude of particle acceleration . `xi`, t and x are in cm, sec, and metre respectively. |
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Answer» We know that the displacement of a particle in a simple harmonic wave is `xi=rsin((2pit)/(T)-(2pix)/(lambda))` Comparing this equation with the equatino `xi=0.25xx10^(-3)sin(500t-025x)`, We have (i) amplitude ` r=0.25xx10^(-3)cm` (ii) `(2pi)/(T)=500 or T=(2pi)/(500)=(pi)/(250)sec` (iii) Angular frequency, `omega=(2pi)/(T)=500rad.//sec` (iv) `(2pi)/(lambda)=-025 or lambda=(2pi)/(0.025) =80pimetres` (v) Amplitude of particle velocity `=` max. velocity `=omega r=(2pi)/(T).r` `=500xx0.25xx10^(-3)=0.125cm//s` (vi) Amplitude of particle acceleratino `=` max. acceleration `=omega^(2)r` `=(500)^(2)xx0.25xx10^(-3)=62.5cm//sec^(2)` |
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