| The linear mass density of the string shown in the figure is \[\mu =1\,g/m.\]. One end of the string is tied to a prong of a tuning fork and the other end carries a block of mass M. The length of the string between the tuning fork and the pulley is\[L=2.0\text{ }m\]. When the tuning fork vibrates, the string resonates with it when mass M is either \[16\text{ }kg\]or\[25\text{ }kg\]. However, standing waves are not observed for any other value of M lying between \[16\text{ }kg\]and \[\text{25 }kg\]. Assume that end A of the string is practically at rest and calculate the frequency of the fork. |
|
A) \[500\,Hz\]
B) \[625\,Hz\]
C) \[900\,Hz\]
D) \[350\,Hz\]
Correct Answer: A
Solution :
| [a] Wave speed \[v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{Mg}{\mu }}\] |
| If the string oscillates with n loops when \[M=16\,kg,\] then there will be \[(n-1)\] loops for\[M=25\text{ }kg\]. |
| \[\because \,\,\,\,f=\frac{n}{2L}\sqrt{\frac{T}{\mu }}\]But frequency is same in both cases. |
| \[\therefore \,\,\,\frac{n}{2L}\sqrt{\frac{16g}{\mu }}=\frac{(n-1)}{2L}\sqrt{\frac{25g}{\mu }}\] |
| \[\therefore \,\,\,\,4n=5(n-1)\Rightarrow n=5\] |
| \[\therefore \,\,\,\,\,f=\frac{5}{2\times 2}\sqrt{\frac{16\times 10}{1\times {{10}^{-3}}}}=\frac{5}{4}\times 4\times {{10}^{2}}=500Hz\] |
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