Sixty complete waves pass a particular point in 4s. If the distance between three successive troughs of the waves is 15cm, calculate the speed of the waves. The period of a wave is 0.02 seconds. Calcu

Frequency is the number of waves per unit time.

$$f = \frac{\text{Number of waves}}{\text{Time}}$$ $$f = \frac{60 \text{ waves}}{4 \text{ s}}$$ $$f = 15 \text{ Hz}$$

The distance between three successive troughs corresponds to two complete wavelengths.

$$2\lambda = 15 \text{ cm}$$ $$\lambda = \frac{15 \text{ cm}}{2}$$ $$\lambda = 7.5 \text{ cm}$$

Convert wavelength to meters:

$$\lambda = 7.5 \times 10^{-2} \text{ m}$$

The speed of a wave is given by the product of its frequency and wavelength.

$$v = f \lambda$$ $$v = (15 \text{ Hz}) \times (7.5 \times 10^{-2} \text{ m})$$ $$v = 1.125 \text{ m/s}$$

The speed of the waves is $\boxed{\text{1.125 m/s}}$.

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The period of a wave is 0.02 seconds. Calculate its wavelength if its speed is 330m/s.

The speed of a wave ($v$) is its wavelength ($\lambda$) divided by its period ($T$).

$$v = \frac{\lambda}{T}$$

Rearrange to solve for wavelength:

$$\lambda = v T$$

Substitute the given values:

$$\lambda = (330 \text{ m/s}) \times (0.02 \text{ s})$$ $$\lambda = 6.6 \text{ m}$$

The wavelength is $\boxed{\text{6.6 m}}$.

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A stationary wave in a spring is fixed at both ends. If the frequency of the wave is 256Hz and the distance N1N2 equals to 1.5m, calculate the speed of the wave.

In a stationary wave, the distance between two successive nodes (N1N2) is half a wavelength ($\frac{\lambda}{2}$).

$$\frac{\lambda}{2} = 1.5 \text{ m}$$ $$\lambda = 2 \times 1.5 \text{ m}$$ $$\lambda = 3.0 \text{ m}$$

The speed of a wave is the product of its frequency and wavelength.

$$v = f \lambda$$ $$v = (256 \text{ Hz}) \times (3.0 \text{ m})$$ $$v = 768 \text{ m/s}$$

The speed of the wave is $\boxed{\text{768 m/s}}$.