The pattern of standing waves formed on a stretched string at two instants of time is shown in the figure. The velocity of the two waves superimposing to form stationary waves is $360 \ m/s$ 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 $A^{\prime}$ and $C^{\prime}$.
$(a)$ Calculate the time at which the second curve is plotted.
$(b)$ Mark nodes and antinodes on the curve.
$(c)$ Calculate the distance between $A^{\prime}$ and $C^{\prime}$.
(N/A) Given: Velocity $v = 360 \ m/s$,Frequency $f = 256 \ Hz$.
Time period $T = \frac{1}{f} = \frac{1}{256} \approx 3.906 \times 10^{-3} \ s$.
$(a)$ At $t = 0$,the string is at its maximum displacement. The second curve (at $t = ?$) represents the string in its equilibrium position (straight line). The time taken for the string to go from maximum displacement to the equilibrium position is $\frac{T}{4}$.
$t = \frac{T}{4} = \frac{3.906 \times 10^{-3}}{4} = 9.765 \times 10^{-4} \ s$.
$(b)$ Nodes are points of zero displacement: $A, B, C, D, E$. Antinodes are points of maximum displacement: $A^{\prime}, C^{\prime}$.
$(c)$ The distance between two consecutive antinodes is equal to the wavelength $\lambda$.
$\lambda = \frac{v}{f} = \frac{360}{256} = 1.40625 \ m \approx 1.41 \ m$.
Time period $T = \frac{1}{f} = \frac{1}{256} \approx 3.906 \times 10^{-3} \ s$.
$(a)$ At $t = 0$,the string is at its maximum displacement. The second curve (at $t = ?$) represents the string in its equilibrium position (straight line). The time taken for the string to go from maximum displacement to the equilibrium position is $\frac{T}{4}$.
$t = \frac{T}{4} = \frac{3.906 \times 10^{-3}}{4} = 9.765 \times 10^{-4} \ s$.
$(b)$ Nodes are points of zero displacement: $A, B, C, D, E$. Antinodes are points of maximum displacement: $A^{\prime}, C^{\prime}$.
$(c)$ The distance between two consecutive antinodes is equal to the wavelength $\lambda$.
$\lambda = \frac{v}{f} = \frac{360}{256} = 1.40625 \ m \approx 1.41 \ m$.
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View SolutionTwo travelling waves produce a standing wave represented by the equation,
${y} = 1.0 \, \text{mm} \cos(1.57 \, \text{cm}^{-1} x) \sin(78.5 \, \text{s}^{-1} t)$
The node closest to the origin in the region ${x} > 0$ will be at ${x} = \dots \, \text{cm}$.
${y} = 1.0 \, \text{mm} \cos(1.57 \, \text{cm}^{-1} x) \sin(78.5 \, \text{s}^{-1} t)$
The node closest to the origin in the region ${x} > 0$ will be at ${x} = \dots \, \text{cm}$.
$A$ string fixed at both the ends forms a standing wave with a node separation of $5 \,cm$. If the velocity of the wave on the string is $2 \,m/s$, then the frequency of vibration of the string is (in $\,Hz$)
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