Answer
(a) Length
(b) $0.21KHz$
(c) $126dB$
Work Step by Step
(a) We know that the thunder stick is considered to be an open pipe as both ends are open. Thus, the frequency of the sound emitted by a thunder stick depends on the length of the air column and we conclude that the frequency is inversely proportional to the length.
(b) We know that
$f=\frac{v}{2L}$
We plug in the known values to obtain:
$f=\frac{343m/s}{2(0.82m)}$
$f=0.21KHz$
(c) We know that
$I=I_{\circ}(10^{\frac{95dB}{10dB}})$
We plug in the known values to obtain:
$I=(10^{-12}W/m^2)(10^{\frac{95dB}{10dB}})$
$\implies I=0.00316W/m^2$
The intensity of $1200$ pairs is given as
$1200(I)=1200(0.00316W/m^2)=3.792W/m^2$
Now $\beta=(10dB)log_{10}(\frac{I}{I_{\circ}})$
We plug in the known values to obtain:
$\beta=(10dB)log_{10}(\frac{3.72W/m^2}{10^{-12}W/m^2})$
$\implies \beta=126dB$
