One end of a taut string of length 3 m along the -axis is fixed at . The speed of the waves in the string is . The other end of the string is vibrating in the -direction so that stationary waves are set up in the string. The possible waveform(s) of these

Question

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 $100 {ms}^{- 1}$ . 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) [2014]

Smart Achievers · Accepted Answer

(1, 3, 4)

There should be a displacement node at $x = 0$ and a displacement antinode at $100 {ms}^{- 1}$ .

Therefore, $y = 0$ at $x = 0$ and $y = \pm A$ at $x = 3 m$ .

Speed of wave, $v = \frac{ω}{k} = 100 {ms}^{- 1}$

Hence options (1), (3) & (4) satisfy the above conditions.

Solution

1 $y (t) = A \sin (\frac{π x}{6}) \cos (\frac{50 π t}{3})$

2 $y (t) = A \sin (\frac{π x}{3}) \cos (\frac{100 π t}{3})$

3 $y (t) = A \sin (\frac{5 π x}{6}) \cos (\frac{250 π t}{3})$

4 $y (t) = A \sin (\frac{5 π x}{2}) \cos (250 π t)$

Ans.
(1, 3, 4)

There should be a displacement node at $x = 0$ and a displacement antinode at $x = 3 m$ .

Therefore, $y = 0$ at $x = 0$ and $y = \pm A$ at $x = 3 m$ .

Speed of wave, $v = \frac{ω}{k} = 100 {ms}^{- 1}$

Hence options (1), (3) & (4) satisfy the above conditions.

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Solution

1 y(t)=Asin(πx6)cos(50πt3)

2 y(t)=Asin(πx3)cos(100πt3)

3 y(t)=Asin(5πx6)cos(250πt3)

4 y(t)=Asin(5πx2)cos(250πt)