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(D) Initial amplitude $A = 10 \, cm$. The total energy of the $SHM$ is $E = \frac{1}{2} k A^2$.At position $x = 5 \, cm$,the velocity $v$ is given by

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$700$

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(D) Initial amplitude $A = 10 \, cm$. The total energy of the $SHM$ is $E = \frac{1}{2} k A^2$.At position $x = 5 \, cm$,the velocity $v$ is given by $v = \omega \sqrt{A^2 - x^2} = \omega \sqrt{10^2 - 5^2} = \omega \sqrt{75}$.Since $\omega^2 = \frac{k}{m}$,we have $v = \sqrt{\frac{k}{m}} \sqrt{75}$.When the velocity is tripled,the new velocity $v' = 3v = 3 \sqrt{\frac{75k}{m}}$.The potential energy at $x = 5 \, cm$ remains unchanged: $U = \frac{1}{2} k (5)^2 = 12.5 k$.The new total energy $E'$ is the sum of the potential energy at $x=5$ and the new kinetic energy $K'$:$E' = \frac{1}{2} k (5)^2 + \frac{1}{2} m (3v)^2 = 12.5 k + \frac{1}{2} m \left( 9 \cdot \frac{75k}{m} \right) = 12.5 k + 337.5 k = 350 k$.Since $E' = \frac{1}{2} k A'^2$,we have $\frac{1}{2} k A'^2 = 350 k \implies A'^2 = 700$.Thus,$A' = \sqrt{700} \, cm$,which implies $x = 700$.

$A$ body is performing simple harmonic motion with an amplitude of $10 \, cm$. The velocity of the body is tripled by an air jet when it is at $5 \, cm$ from its mean position. The new amplitude of vibration is $\sqrt{x} \, cm$. The value of $x$ is . . . . . . .

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