- Part (a): Frequency \(f = 1.25 \text{ Hz}\), Period \(T = 0.8 \text{ s}\), Velocity \(v = 2.5 \text{ cm/s}\).
- Part (b): Frequency \(f = 1.5 \times 10^7 \text{ Hz}\), Periodic time \(T \approx 6.67 \times 10^{-8} \text{ s}\), Cycles \(N = 2.7 \times 10^{11}\).
- Part (c): Wavelength \(\lambda = \frac{50}{9} \text{ cm}\), Velocity \(v = 1 \text{ m/s}\).
- \(\boxed{a) 1.25 \text{ Hz}, 0.8 \text{ s}, 2.5 \text{ cm/s}; b) 1.5 \times 10^7 \text{ Hz}, 6.67 \times 10^{-8} \text{ s}, 2.7 \times 10^{11} \text{ cycles}; c) 1 \text{ m/s}}\)
Explanation
- Part (a): Calculating Frequency, Period, and Velocity
AFor a wave that makes 10 cycles in 8 seconds, the frequency \(f\) is calculated as:
\[f = \frac{\text{number of cycles}}{\text{time}} = \frac{10}{8} = 1.25 \text{ Hz}\]
The period \(T\) is the reciprocal of the frequency:
\[T = \frac{1}{f} = \frac{1}{1.25} = 0.8 \text{ s}\]
Given the wavelength \(\lambda = 2 \text{ cm} = 0.02 \text{ m}\), the velocity \(v\) is:
\[v = f \times \lambda = 1.25 \text{ Hz} \times 0.02 \text{ m} = 0.025 \text{ m/s}\]
(Or in cm/s: \(v = 1.25 \text{ Hz} \times 2 \text{ cm} = 2.5 \text{ cm/s}\)).
- Part (b): Calculating Frequency, Periodic Time, and Total Cycles
BGiven \(\lambda = 2000 \text{ cm} = 20 \text{ m}\) and \(v = 3.0 \times 10^8 \text{ m/s}\):
The frequency \(f\) is:
\[f = \frac{v}{\lambda} = \frac{3.0 \times 10^8 \text{ m/s}}{20 \text{ m}} = 1.5 \times 10^7 \text{ Hz}\]
The periodic time \(T\) is:
\[T = \frac{1}{f} = \frac{1}{1.5 \times 10^7 \text{ Hz}} \approx 6.67 \times 10^{-8} \text{ s}\]
The number of cycles \(N\) in 5 hours (\(t = 5 \times 3600 \text{ s} = 18000 \text{ s}\)):
\[N = f \times t = 1.5 \times 10^7 \text{ Hz} \times 18000 \text{ s} = 2.7 \times 10^{11} \text{ cycles}\]
- Part (c): Determining the Velocity of the Wave
CFor 10 consecutive troughs, the distance between the first and the tenth trough is \(9\lambda = 50 \text{ cm}\).
Thus, the wavelength \(\lambda\) is:
\[\lambda = \frac{50}{9} \text{ cm} \approx 5.56 \text{ cm} = 0.0556 \text{ m}\]
With a frequency \(f = 18 \text{ Hz}\), the velocity \(v\) is:
\[v = f \times \lambda = 18 \text{ Hz} \times \frac{50}{9} \text{ cm} = 100 \text{ cm/s} = 1 \text{ m/s}\]
Examples
Understanding wave properties is fundamental in telecommunications and oceanography. For example, by measuring the distance between successive waves (wavelength) and the time it takes for them to pass a point (frequency), engineers can design radio antennas to correctly capture signal frequencies or marine researchers can estimate the speed of ocean waves to predict tide changes. These calculations are essential for ensuring that technology and infrastructure operate efficiently in real-world conditions.