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The vibrations of a string length 60 cm fixed at both ends are represented by the equation – y = 4 sin (πx/15) cos (96 πt) Where x and y are in cm and t in seconds. (i) What is the maximum displacement of a point at x = 5 cm? (ii) Where are the nodes located along the string? (iii) What is the velocity of the particle at x = 7.5 cm at t = 0.25 sec.? (iv) Write down the equations of the component waves whose superposition gives the above wave.

The vibrations of a string length 60 cm fixed at both ends are represented by the equation –
y = 4 sin (πx/15) cos (96 πt)
Where x and y are in cm and t in seconds.
(i)  What is the maximum displacement of a point at x = 5 cm?
(ii) Where are the nodes located along the string?
(iii) What is the velocity of the particle at x = 7.5 cm at t = 0.25 sec.?
(iv) Write down the equations of the component waves whose superposition gives the above wave.

Grade:11

2 Answers

Aditi Chauhan
askIITians Faculty 396 Points
7 years ago
Hello Student,
Please find the answer to your question
(i) Here amplitude, A = sin (π x/15)
At x = 5m
A = 4 sin (π x 5/15) = 4 x 0.866 = 3.46 cm
(ii) Nodes are the position where A = 0
∴ sin (π x/15) = 0 = sin n π ∴ c = 15 n
Where n = 0, 1, 2 x = 15 cm, 30 cm, 60 cm, . . . . . . . . . . . . . .
(iii) y = 4 sin (π x/15) cos (96 π t)
v = dy/dt = 4 sin (π x/15) [- 96 π sin (96 π t)]
At x = 7.5 cm, t = 0.25 cm
v = 4 sin (π x 7.5/15) [- 96 π sin (96 π x 0.25) ]
= 4 sin (π/2) [-96 π sin (24 π)] = 0
(iv) y = 4 sin (π x/15) cos [96 π t]
= 2[ 2 sin (π x/15) cos (96 π t)]
= 2[ sin (96 π t + π r/15) – sin (96 π t – π x/15)]
= 2 sin (96 π t + π x/15) – 2 sin (96 π t – π x/15)
= y1 + y2
Where y1 = 2 sin (96 π t + π x/15)
And y2 = -2 sin (96 π t – π x/15)

Thanks
Aditi Chauhan
askIITians Faculty
Kushagra Madhukar
askIITians Faculty 629 Points
one year ago
Dear student,
Please find the answer to your question.
 
Please find the answer to your question
 
(i) Here amplitude, A = sin (π x/15)
At x = 5m
A = 4 sin (π x 5/15) = 4 x 0.866 = 3.46 cm
 
(ii) Nodes are the position where A = 0
∴ sin (π x/15) = 0 = sin n π ∴ c = 15 n
Where n = 0, 1, 2 x = 15 cm, 30 cm, 60 cm, . . . . . . . . . . . . . .
 
(iii) y = 4 sin (π x/15) cos (96 π t)
v = dy/dt = 4 sin (π x/15) [- 96 π sin (96 π t)]
At x = 7.5 cm, t = 0.25 cm
v = 4 sin (π x 7.5/15) [- 96 π sin (96 π x 0.25) ]
= 4 sin (π/2) [-96 π sin (24 π)] = 0
 
(iv) y = 4 sin (π x/15) cos [96 π t]
= 2[ 2 sin (π x/15) cos (96 π t)]
= 2[ sin (96 π t + π r/15) – sin (96 π t – π x/15)]
= 2 sin (96 π t + π x/15) – 2 sin (96 π t – π x/15)
= y1 + y2
Where y1 = 2 sin (96 π t + π x/15)
And y2 = -2 sin (96 π t – π x/15)
 
Thanks and regards,
Kushagra

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