Kevin Nash
Last Activity: 10 Years ago
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)