Considering the wave described in answer the following questions;

(a) Are all the points in the wire oscillating at the same values of

(i) frequency,

(ii) phase,

(b) Calculate the amplitude of a point 0.4 m away from one end?

Asked by Pragya Singh | 1 year ago |  101

##### Solution :-

(a) As the wire is clamped at both its ends, the ends behave as nodes and the whole wire vibrates in one segment. Thus,

(i) Except at the ends which have zero frequency, all the particles in the wire oscillate with the same frequency.

(ii) All the particles in the wire lie in one segment, thus they all have the same phase. Except for the nodes.

(iii) Amplitude, however, is different for different points.

(b) Given equation,

y (x, t) =0.06sin$$(\dfrac{2\pi}{3}x)cos(120\pi t)$$

For x = 0.4m and t =0

Amplitude = displacement =0.06sin$$(\dfrac{2\pi}{3}x)cos0$$

= 0.06sin$$(\dfrac{2\pi}{3}\times 0.4)1$$

= 0.044 m

Answered by Pragya Singh | 1 year ago

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