In case of a stationary wave pattern which of the following statement is CORRECT?
In case of a stationary wave pattern which of the following statement is CORRECT?
The distance between the consecutive nodes is equal to the wavelength.
In a pipe at both ends only even harmonics are present in an air column.
In a pipe closed at one end, all harmonics are present in an air column.
In case of a stretched string when vibrated, frequency of first overtone is same as second harmonic.
Solution
Frequency of first harmonic $=\mathrm{n}=\frac{\mathrm{v}}{\lambda}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$ Frequency of second harmonic $=2 \mathrm{n}=\frac{1}{l} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}.... (i)$
For the first overtone, $\lambda=l$
$\therefore \quad$ Frequency of first overtone $\mathrm{n}_1=\frac{1}{l} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}.... (ii)$
Comparing (i) and (ii),
$\mathrm{n}_1=2 \mathrm{n}$