JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Waves & Sound

One end of a taut string of length

3 m

along the

x

axis is fixed at

x = 0

. The speed of the waves in the string is

100 m s^{- 1}

. The other end of the string is vibrating in the

y

direction so that stationary waves are set up in the string. The possible waveform(s) of these stationary waves is(are)

Answer: (A), (C), (D)

Solution

For stationary waves

y = A sin (k x) cos (ω t)

, wave speed

v = \frac{ω}{k}

. One end fixed at

x = 0

(node), other end at

x = 3

m vibrating (antinode).

Check each option:
(A)

v = \frac{50 π /3}{π /6} = 100

m/s ✓. At

x = 3

sin (\frac{π \times 3}{6}) = sin (\frac{π}{2}) = 1

(antinode) ✓
(B)

v = \frac{100 π /3}{π /3} = 100

m/s ✓. At

x = 3

sin (π) = 0

(node) ✗
(C)

v = \frac{250 π /3}{5 π /6} = 100

m/s ✓. At

x = 3

sin (\frac{5 π}{2}) = 1

(antinode) ✓
(D)

v = \frac{250 π}{5 π /2} = 100

m/s ✓. At

x = 3

sin (\frac{15 π}{2}) = sin (\frac{π}{2}) = 1

(antinode) ✓

Options (A), (C), and (D) are correct.

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · One or more correct options