A) amplitude of motion is 4a
B) time period of oscillations is \[6\tau \]
C) amplitude of motion is 3a
D) time period of oscillations is \[8\tau \]
Correct Answer: B
Solution :
\[\cos (\omega \tau )=\frac{A-a}{A}\] \[\cos (2\omega \tau )=\frac{A-3a}{A}\] \[\Rightarrow \]\[\frac{A-3a}{A}=2{{\left( \frac{A-a}{A} \right)}^{2}}-1\] \[\Rightarrow \]\[1-\frac{3a}{A}=2+2{{\left( \frac{a}{A} \right)}^{2}}-\frac{4a}{A}-1\] \[\left( \frac{a}{A} \right)={{\left( \frac{a}{A} \right)}^{2}}A=2a\] \[\cos (\omega \tau )=\frac{1}{2}\Rightarrow \omega \tau =\frac{\pi }{3}\Rightarrow \frac{2\pi }{T}.\tau =\frac{\pi }{3}\Rightarrow T=6\tau \]You need to login to perform this action.
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