Q.A stationary wave on a stretched string is recorded at two instants of time. At the first instant () the string is sinusoidal: with the transverse displacement on the vertical axis and position (in metres) on the horizontal axis, the string touches the axis (zero displacement) at five equally spaced points labelled, in order along , , , , and . Between consecutive touch points the string bulges alternately upward and downward — a crest in the - loop, a trough in the - loop, a crest in the - loop and a trough in the - loop. The antinode (point of maximum displacement) in the first loop is labelled and the antinode in the third loop (between and ) is labelled . At the second instant the string is completely straight, with zero displacement everywhere. The two waves that superpose to form this stationary wave travel at 360 m/s and each has a frequency of 256 Hz.
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Start your 14-day free trial to unlock the full solution →With m/s and Hz, the wavelength is m and the period is s. The string is straight when the time factor , first at s. Nodes are the fixed axis-crossings ; antinodes are the loop-centres (and the other two). and are antinodes one full wavelength apart, so m.
Given
m/s, Hz.
(a) Time of the straight profile
A stationary wave is . The string is straight everywhere when the time factor vanishes, , first at , i.e. :
(b) Nodes and antinodes
Nodes are the permanently stationary points where — the axis-crossings . Antinodes are the points of maximum displacement, mid-way between consecutive nodes — the centres of the four loops (the crests/troughs), which include and . …
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