A source of sound produces waves in air of wavelength 1.65m. If the speed of sound in air is 330ms$^{-1}$, the period of vibration in air is?

Analysis

To solve this problem, we need to relate the speed of sound, the wavelength, and the period of vibration.

Concept

The relationship between the speed of a wave ($v$), its frequency ($f$), and its wavelength ($\lambda$) is given by the formula:

We also know that frequency ($f$) is the reciprocal of the period ($T$):

By substituting the second equation into the first, we get the relationship between speed, wavelength, and period:

Rearranging this to solve for the period ($T$):

Step-by-Step Solution

1. Identify the given values: • Wavelength ($\lambda$) = $1.65 \, \text{m}$ • Speed of sound ($v$) = $330 \, \text{ms}^{-1}$

2. Apply the formula: $$T = \frac{1.65}{330}$$

3. Calculate the result: • To make the division easier, you can write $1.65$ as $\frac{165}{100}$: $$T = \frac{165}{100 \times 330}$$ • Since $165$ is half of $330$ ($165 \times 2 = 330$): $$T = \frac{1}{100 \times 2}$$ $$T = \frac{1}{200}$$ $$T = 0.005 \, \text{s}$$

Conclusion

The period of vibration is $0.005$ seconds. This matches option B.

Exam Tip

Always check your units before calculating. In JAMB Physics, speed is usually in $\text{ms}^{-1}$ and wavelength in meters ($\text{m}$), which gives the period in seconds ($\text{s}$). If you ever forget the formula, remember that Speed = Distance / Time. In one period ($T$), a wave travels exactly one wavelength ($\lambda$). Therefore, $v = \lambda / T$.