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.

Question

Aditi Chauhan · Answer

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 · Answer

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) = y 1 + y 2 Where y 1 = 2 sin (96 π t + π x/15) And y 2 = -2 sin (96 π t – π x/15) Thanks and regards, Kushagra

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