Waves Test - 61

$$s=330$$

$$h=1.5cm=1.5/100=0.015$$

$$s×h$$

$$330×0.015$$

$$=4.95 Hz$$

it Will not be audible as it less than $$20Hz$$

Hence,

option $$(C)$$ is correct answer.

$$\begin{array}{l} Given, \\ Diplacement\, \, function\, \, of\, \, the\, \, particle\, \, is, \\ x\left( t \right) =A\cos  \left( { \omega t+\phi  } \right)  \\ x\left( 0 \right) =1\, \, cm\, \, and\, \, \, v\left( 0 \right) \, \, =\pi \, \, \dfrac { { cm } }{ { \sec   } }  \\ differentiating\, \, w.r.\, \, to\, \, t, \\ v\left( t \right) =-A\left( \omega  \right) \sin  \left( { \omega t+\varphi  } \right) ..........\left( 1 \right)  \\ At\, \, t=0, \\ x\left( 0 \right) =A\omega \sin  \left( \phi  \right) ..........\left( 2 \right)  \\ v\left( 0 \right) =-A\omega \sin  \left( \phi  \right) ............\left( 3 \right)  \\ Dividing\, \, \left( 3 \right) \, \, by\, \, \left( 2 \right) , \\ \dfrac { { v\left( 0 \right)  } }{ { x\left( 0 \right)  } } =-\omega \tan  \phi  \\ \Rightarrow \pi =-\omega \tan  \phi  \\ \Rightarrow \tan  \phi =-1 \\ i.e\, \, \, \phi =\dfrac { { 3\pi  } }{ 4 } \, \, or\, \, \dfrac { { 7\pi  } }{ 4 }  \\ \sin  ce\, \, ,at\, \, \, t=0 \\ x\left( 0 \right) =A\cos  \left( \phi  \right) =1 \\ \Rightarrow \cos  \phi \, \, \, is\, \, \, positive. \\ Therefore\, \, ,\phi =\dfrac { { 7\pi  } }{ 4 }  \\ \therefore A\cos  \left( { \dfrac { { 7\pi  } }{ 4 }  } \right) =1 \\ \Rightarrow A=\sqrt { 2 } \, \, cm \end{array}$$