Wave motionHardBloom L2
Question
The vibration of a string fixed at both ends are described by Y= 2 sin(πx) sin(100πt) where Y is in mm,x is in cm,t in sec then
Options
A.Maximum displacement of the particle at x = 1/6 cm would be 1 mm.
B.velocity of the particle at x = 1/6 cm at time t = 1/600 sec will be 157 √3 mm/s
C.If the length of the string be 10 cm, number of loops in it would be 5
D.None of these
Solution
Sol. (A) at x = $\frac{1}{6}$cm
y = 2 sin$\left( \frac{\pi}{6} \right)$sin(100πt) = (1mm) sin(100πt)
(B) vp = $\frac{\partial y}{\partial t}$ = 2 × 100π sin(πx) cos(100πt)
Put x = $\frac{1}{6}$& t = $\frac{1}{600}$
(C) K = π = $\frac{2\pi}{\lambda}$
⇒λ = 2cm
L = 10cm = 10 × $\frac{\lambda}{2}$
$\frac{\lambda}{2}$→ = 1 loop
⇒ Total 10 loops
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