Figure shows a string stretched by a block going over a pulley. The string vibrates in its tenth harmonic in…
Figure shows a string stretched by a block going over a pulley. The string vibrates in its tenth harmonic in unison with a particular tuning fork.
When a beaker containing water is brought under the block, so that the block is completely dipped into the beaker, the string vibrates in its eleventh harmonic. Find the density of the material of the block.
Solution
Sol. According to the question,
Frequency, $f=\frac{10}{2l}\sqrt{\frac{T}{\alpha}}=\frac{10}{2l}\sqrt{\frac{mg}{\alpha}}$ ...(i)
The free body diagram of the given situation is shown below.
$\therefore\;T'+F_B=mg$
$T'=mg-F_B=mg-\rho_w\frac{m}{\sigma}g$
$=mg\left(1-\frac{\rho_w}{\sigma}\right)$
$\therefore\;f=\frac{11}{2l}\sqrt{\frac{T'}{\alpha}}=\frac{11}{2l}\sqrt{\frac{mg\left(1-\frac{\rho_w}{\sigma}\right)}{\alpha}}$ ...(ii)
Equating Eqs. (i) and (ii), we get
$\frac{10}{2l}\sqrt{\frac{mg}{\alpha}}=\frac{11}{2l}\sqrt{\frac{mg\left(1-\frac{\rho_w}{\sigma}\right)}{\alpha}}$
$\Rightarrow\;10=11\sqrt{1-\frac{\rho_w}{\sigma}}$
$\Rightarrow\;1-\frac{\rho_w}{\sigma}=\left(\frac{10}{11}\right)^2=\frac{100}{121}$
$\Rightarrow\;\frac{\rho_w}{\sigma}=1-\frac{100}{121}=\frac{21}{121}$
$\Rightarrow\;\sigma=\frac{121}{21}\rho_w=\frac{121}{21}\,{\rm g/cm^3}=5.8\,{\rm g/cm^3}$
(where, $\sigma$ is density of material of block and $\rho_w$ is density of water.)
Answer: 5.8 g/cm$^{3}$