A) \[\lambda =18\,cm\]
B) \[v=4\,m/s\]
C) \[a=0.4\,m\]
D) \[f=50\,Hz\]
Correct Answer: A
Solution :
Key Idea: Compare the given wave equation with standard one. The given equation be written as \[y=4\,\sin \left[ \pi \left( \frac{t}{5}-\frac{x}{9}+\frac{1}{6} \right) \right]\] ... (i) The standard wave equation can be written as \[y=a\,\sin \,(\omega t-kx+\phi )\] or \[y=a\sin \left( \frac{2\pi }{T}t-\frac{2\pi }{\lambda }.\frac{x}{1}+\phi \right)\] ? (ii) Equating Eqs. (i) and (ii), we get Amplitude, \[a=4\,cm=0.04\,m\] Frequency, \[f=\frac{1}{T}=\frac{1}{10}Hz=0.1\,Hz\] Wavelength, \[\lambda =2\times 9=18\,cm\] and velocity, \[v=f\,\lambda =0.1\times 18=1.8\,cm/s\]
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