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In a standing wave experiment , a 1.2 - kg horizontal rope is fixed in place at its two ends ( x = 0 and x = 2.0 m) and made to oscillate up and down in the fundamental mode , at frequency of 5.0 Hz. At t = 0 , the point at x = 1.0 m has zero displacement and is moving upward in the positive direction of y - axis with a transverse velocity 3.14 m//s. What is the correct expression of the standing wave equation ?

Answer» <html><body><p>`(0.1) sin ( pi//2) x sin (10 pi) t`<br/>`(0.1) sin (pi) x sin (10 pi) t`<br/>`(0.05) sin (pi//2) x cos (10 pi) t`<br/>`(0.04) sin (pi//2) x sin (10 pi) t`</p>Solution :`mu = (1.2)/(2) = 0.6 kg//m` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a> = 5 Hz` <br/> `lambda = 2 l = <a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a> m` <br/> `V = n lambda = 5 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 4 = 20 m//s` <br/> Using `v = sqrt((T)/(mu))` <br/> `T = 20^(2) xx 0.6 = 240 N` <br/> `((delta y)/( delta t))_(max) = 3.14 m//s` <br/> `( 2 A)omega = 3.14` <br/> Amplitude `2A = (3.14)/(2 xx (3.14) xx 5) = 0.1 m` <br/> <a href="https://interviewquestions.tuteehub.com/tag/equation-974081" style="font-weight:bold;" target="_blank" title="Click to know more about EQUATION">EQUATION</a> of standing wave is <br/> `y = (0.1) sin (pi)/(2) s sin (10 pi) t`</body></html>


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