Question

1)

The pattern of standing waves formed on a stretched string at two instants of time are shown in figure. The velocity of two waves superimposing to form stationary waves is $360m/s$ and their frequencies are $256Hz$.

A) Calculate the time at which the second curve is plotted



2) The pattern of standing waves formed on a stretched string at two instants of time are shown in figure. The velocity of two waves superimposing to form stationary waves is $360m/s$ and their frequencies are $256Hz$.

B) Mark nodes and antinodes on the curve



3)

The pattern of standing waves formed on a stretched string at two instants of time are shown in figure. The velocity of two waves superimposing to form stationary waves is $360m/s$ and their frequencies are $256Hz$.

C) Calculate the distance between $A\text{'}$ and $C\text{'}$

Answer 1

1)



Given, frequency of the wave,

$\nu =256Hz$

Time period $T=\frac{1}{\nu }=\frac{1}{256}s$

$=3.9\times {10}^{-3}$ second

In stationary wave, a particle passes through its mean position after every $\frac{T}{4}$ time,

$t=\frac{T}{4}=\frac{3.9\times {10}^{-3}}{4}s$

$=9.8\times {10}^{-4}second$

Final Answer: $9.8\times {10}^{-4}second$


2)


When the particle is at its maximum displacement it shows antinodes. And particle is at its mean position is represents nodes.

So, Nodes are at position $A,B,C,D,E$ and Antinodes are at position $A\text{'},C\text{'}$.

Final Answer: Nodes are $A,B,C,D,E$ and Antinodes are $A\text{'},C\text{'}$.

3)


Given, frequency of the wave,

$\nu =256Hz$

velocity of stationary waves $=360m/s$

The point $A^{\prime }$ and $C^{\prime }$ are at maximum displacement.

Separation between $A^{\prime }$ and $C^{\prime }=$ wavelength$\left(\lambda \right)$

$\lambda =\frac{v}{\nu }=\frac{360}{256}=1.41m$

Final Answer: $1.41m$