0

Guest

Courses New

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
8 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 628 Points
2 years 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

Think You Can Provide A Better Answer ?

Other Related Questions on Wave Motion

View all Questions »

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More