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Sol. . Y = (1 mm) sin π [x/2cm – t/0.01 sec] a) T = 2 0.01 = 0.02 sec = 20 ms λ = 2 * 2 = 4 cm b) v = dy/dt = d/dt [sin 2π (x/4 – t/0.02)] = –cos2π {x/4) – (t/0.02)} * 1/(0.02) ⇒ v = –50 cos 2π {(x/4) – (t/0.02)} at x = 1 and t = 0.01 sec, v = –50 cos 2* [(1/4) – (1/2)] = 0 c) i) at x = 3 cm, t = 0.01 sec v = –50 cos 2π (3/4 – ½) = 0 ii) at x = 5 cm, t = 0.01 sec, v = 0 (putting the values) iii) at x = 7 cm, t = 0.01 sec, v = 0 at x = 1 cm and t = 0.011 sec v = –50 cos 2π {(1/4) – (0.011/0.02)} = –50 cos (3/5) = –9.7 cm/sec (similarly the other two can be calculated)
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