Question

The vibrations of a string of length 60 cm fixed at both ends are represented by the equation y=4sin((pix)/15) cos (96 pi t), where x and y are in cm and t in seconds. (a)What is the maximum displacement of a point at x = 5cm? (b)Where are the nodes located along the string? (c)What is the velocity of the particle at x=7.5cm and t=0.25s? (d)Write down the equations of the component waves whose superposition gives the above wave.

Answer 1

Given:

String of length 60 cm fixed at both ends.

Represented by the equation y=4sin((pix)/15) cos (96 pi t).

To Find:

1. The maximum displacement of a point at x = 5cm.
2. The nodes located along the string.
3. The velocity of the particle at x=7.5cm and t=0.25s.
4. The equations of the component waves whose superposition gives the above wave.

Solution:

• y = 4 sin (πx/15) cos(96πt)

Therefore,

1. The maximum displacement at x =5cm,

• y = 4 sin(5π/15)cos(96πt)
• y(x=5) = 4 sin(π/3)cos(96πt)

For maximum y, cos should be maximum.

Therefore,

• Ymax = 4 sin(π/3)x 1
• Ymax = 2√3 cm.

    2. At Nodes y = 0

• 4 sin(πx/15) = 0
• πx/15 = nπ
• x = 15n cm , n = 0, 1,2,...
• x = {0, 15cm , 30cm ,45cm ,60cm }

    3. Velcoity of particle at x = 7.5cm and t = 0.25s

• dy/dt = 4sin(πx/15)sin(96t) x 96
• V = 384 x π x sin(π/2) sin(24π)
• V = 0

    4. Component waves whose superposition gives the above wave.

• We know 2SinA cosB = Sin(A + B) + Sin(A-B)

• y =  4 sin (πx/15) cos(96πt)
• y = 2 sin(πx/15 + 96πt) + 2sin(πx/t - 96πt)

The maximum displacement of a point at x = 5cm, is Ymax = 2√3cm.

The nodes located at 0, 15cm , 30cm ,45cm ,60cm lengths of the string.

The velocity of the particle at x=7.5cm and t=0.25s is 0.

The equations of the component waves whose superposition gives the above wave is  2 sin(πx/15 + 96πt) + 2sin(πx/t - 96πt).