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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