Question

You are listening to an $A$ note played on a violin string. Let the subscript $s$ refer to the violin string and $a$ refer to the air. Then:

• A. $f_s=f_a$ but ${\lambda }_s\ne {\lambda }_a$
• B. $f_s=f_a$ and ${\lambda }_s={\lambda }_a$
• C. ${\lambda }_s={\lambda }_a$ but $f_s\ne f_a$
• D. ${\lambda }_s\ne {\lambda }_a$ and $f_s\ne f_a$

Answer 1

The correct option is A $f_s=f_a$ but ${\lambda }_s\ne {\lambda }_a$

Frequencies of air and string is same because the source is same. Air and the string are two different medium so there velocities will also different. Hence the wavelength has to be different.