Ncert Exercises & Solutions

The patterns of standing waves formed on a stretched string at two instant of time are shown in the figure. The velocity of two waves superimposing to form stationary waves is 360 ${ms}^{- 1}$ and their frequency is 256 Hz.

1. Calculate the time at which the second curve is plotted.

2. Mark nodes and antinodes on the curve.

3. Calculate the distance between A' and C'.

Hint: Use the concept of standing waves.

Given, the frequency of the wave,

ν

= 256 Hz

Step 1: Find the time of passing through the mean position.

Time period,

T = \frac{1}{ν} = \frac{1}{256} s = 3.9 \times 10^{- 3} s

1. Time taken to pass through mean position is:

t = \frac{T}{4} = \frac{1}{40} = \frac{3.9 \times 10^{- 3}}{4} s = 9.8 \times 10^{- 4} s

Step 2: Find the positions of nodes and anti-nodes.

2. Nodes are A, B, C, D, E (i.e., zero displacements)

Antinodes are A', C' (i.e., maximum displacement)

Step 3: Find the wavelength.

3. It is clear from the diagram A' and C' are consecutive antinodes, hence,

separation = wavelength (

λ

)

= \frac{v}{ν} = \frac{360}{256} = 1.41 m [∴ v = νλ]

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