The pattern of standing waves formed on a stretched string at two instants of time are shown in Fig. The velocity of two waves superimposing to form stationary waves is 360 ms–1 and their frequencies are 256 Hz

(a) Calculate the time at which the second curve is plotted.

(b) Mark nodes and antinodes on the curve.

(c) Calculate the distance between A′ and C′ .

3 Views|Posted 7 months ago
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7 months ago

This is a short answer type question as classified in NCERT Exemplar

Given frequency of the wave v=256Hz

          T= 1 v = 1 256 = 3.9 * 10 3 s

(a) time taken to pass through mean position is

t=T/4=1/40 =3.9 * 10 - 3 4 s = 9.8 * 10-4s

(b) nodes are A, B, C, D, E (having zero displacement)

antinodes are A' and C' (having maximum displacement)

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Physics NCERT Exemplar Solutions Class 11th Chapter Fifteen 2025

Physics NCERT Exemplar Solutions Class 11th Chapter Fifteen 2025

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