Chapter 26, Problem 49A

(a)

To determine

To show that period of a keyboard tone with frequency 256 Hz is 0.00391 s.

Answer to Problem 49A

The period of a keyboard tone with frequency 256 Hz is 0.00391 s.

Explanation of Solution

Given:

Frequency ( $\mathrm{f}$ ) of keyboard tone = 256 Hz

Formula used:

Time period of vibration ( $\mathrm{t}$ ) = $\frac{1}{frequency(f)ofthetone}$

Calculation:

Put the given values and we get;

Time period of vibration ( $\mathrm{t}$ ) = $\frac{1}{256Hz}=0.00391s$

Conclusion:

(b)

To determine

To show that the wavelength of sound wave in air is 1.33 m for the frequency 256 Hz.

Answer to Problem 49A

The wavelength of sound wave in air is 1.33 m for the frequency 256 Hz.

Explanation of Solution

Given:

Speed of sound wave ( $\mathrm{s}$ ) = 340 ${\mathrm{ms}}^{-1}$

Frequency ( $\mathrm{f}$ ) of keyboard tone = 256 Hz

Formula used:

  $\mathrm{Wavelength}(\lambda )ofsoundwave=\frac{speed(s)ofsoundwave}{frequency(f)}$

Calculation:

Put the values and we get;

  $\mathrm{Wavelength}(\lambda )ofsoundwave=\frac{340m{s}^{-1}}{256Hz}=1.33m$

  $\mathrm{Wavelength}(\lambda )ofsoundwave=1.33m$

Conclusion:

Thus, the wavelength of sound wave in air is 1.33 m for the frequency 256 Hz.