A guitar string vibrates at a frequency of 330 Hz with a wavelength 1.40 m. The frequency and wavelength of this sound in air (20°C) as it reaches our ears is

(a) same frequency, same wavelength.

(b) higher frequency, same wavelength.

(c) lower frequency, same wavelength.

(d) same frequency, longer wavelength.

(e) same frequency, shorter wavelength.

Wavelengthcan be defined as the distance between two successive crests or troughs of a wave.It is measured in lengths such as meters, centimeters, millimeters, nanometers, and others.

Frequency is the speed of vibration, and this determines the pitch of the sound. It is measured in Hertz.

As the string oscillates, it causes the air to vibrate at the same frequency. Therefore, the sound wave will have the same frequency as the guitar string, so answers (b) and (c) are incorrect. The speed of sound in the air at \({20^ \circ }\;{\rm{C}}\) temperature is \(343\;{{\rm{m}} \mathord{\left/

{\vphantom {{\rm{m}} {\rm{s}}}} \right.

\\{\rm{s}}}\). Sound speed in the string is the product of the wavelength and frequency \(462\;{{\rm{m}} \mathord{\left/

{\vphantom {{\rm{m}} {\rm{s}}}} \right.

\\{\rm{s}}}\), so the sound waves in the air have a shorter wavelength than the waves on the string.