To analyze the standing wave pattern described by the equation \( y(x, t) = 0.040 \sin(5\pi x) \cos(40\pi t) \), we can break down the problem into several parts. This equation represents a standing wave formed by the interference of two traveling waves. Let's tackle each part of your question step by step.
Finding Node Locations
Nodes in a standing wave occur where the displacement is always zero. This happens when the sine component of the wave equation is zero. Therefore, we need to solve:
1. For Nodes:
\( \sin(5\pi x) = 0 \)
The sine function is zero at integer multiples of \( \pi \). Thus, we can set:
\( 5\pi x = n\pi \) where \( n \) is an integer.
Solving for \( x \), we get:
\( x = \frac{n}{5} \)
Now, we can find the smallest, second smallest, and third smallest values of \( x \) by substituting \( n = 0, 1, 2 \):
- (a) Smallest node: \( n = 0 \) gives \( x = 0 \) m
- (b) Second smallest node: \( n = 1 \) gives \( x = 0.2 \) m
- (c) Third smallest node: \( n = 2 \) gives \( x = 0.4 \) m
Period of Oscillatory Motion
Next, we need to determine the period of the oscillatory motion of any non-node point. The period \( T \) can be found from the cosine term:
2. For Period:
The cosine function \( \cos(40\pi t) \) has a frequency \( f \) given by:
\( f = \frac{40\pi}{2\pi} = 20 \) Hz
Thus, the period \( T \) is the reciprocal of the frequency:
\( T = \frac{1}{f} = \frac{1}{20} = 0.05 \) seconds
Speed and Amplitude of Traveling Waves
Now, let's find the speed and amplitude of the two traveling waves that interfere to create this standing wave.
3. For Speed:
The wave number \( k \) is given by the coefficient of \( x \) in the sine function:
\( k = 5\pi \) m-1
The speed \( v \) of the wave can be calculated using the relationship \( v = f \lambda \), where \( \lambda \) is the wavelength. The wavelength \( \lambda \) can be found from \( k \):
\( \lambda = \frac{2\pi}{k} = \frac{2\pi}{5\pi} = \frac{2}{5} = 0.4 \) m
Now, substituting into the speed formula:
\( v = 20 \times 0.4 = 8 \) m/s
4. For Amplitude:
The amplitude of the standing wave is given directly in the equation as \( 0.040 \) m. Each traveling wave has an amplitude of half of the standing wave's amplitude:
Amplitude of each wave = \( \frac{0.040}{2} = 0.020 \) m
Times of Zero Transverse Velocity
Finally, we need to find the times when all points on the string have zero transverse velocity. The transverse velocity is given by the time derivative of \( y(x, t) \):
5. For Zero Transverse Velocity:
Taking the derivative:
\( v_y = \frac{\partial y}{\partial t} = 0.040 \sin(5\pi x)(-40\pi \sin(40\pi t)) \)
Setting this equal to zero gives us two conditions:
- When \( \sin(5\pi x) = 0 \) (which we already solved for nodes)
- When \( \sin(40\pi t) = 0 \)
For \( \sin(40\pi t) = 0 \), we have:
\( 40\pi t = m\pi \) where \( m \) is an integer.
Solving for \( t \):
\( t = \frac{m}{40} \)
Now, substituting \( m = 0, 1, 2 \):
- (g) First time: \( t = 0 \) seconds
- (h) Second time: \( t = \frac{1}{40} = 0.025 \) seconds
- (i) Third time: \( t = \frac{2}{40} = 0.050 \) seconds
In summary, we have identified the locations of the nodes, calculated the period of oscillation, determined the speed and amplitude of the traveling waves, and found the times when all points on the string have zero transverse velocity. Each of these steps builds a comprehensive understanding of the standing wave behavior on the string.
