A) 5 cm
B) 7.5 cm
C) 10 cm
D) 12.5 cm
Correct Answer: A
Solution :
Velocity of SHM \[v=\omega \sqrt{{{a}^{2}}-{{y}^{2}}}\] \[\Rightarrow \] \[9={{\omega }^{2}}[{{a}^{2}}-16]\] ... (i) Similarly, \[16={{\omega }^{2}}[{{a}^{2}}-9]\] ... (ii) Dividing Eq. (ii) by Eq. (i), we get \[\frac{16}{9}=\frac{[{{a}^{2}}-9]}{[{{a}^{2}}-16]}\] \[\Rightarrow \] \[{{a}^{2}}=25\] or a \[=5\text{ }cm\]
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