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A body with speed v is moving along a straight line. At the same time it is at distance from a fixed point on the line, the speed is given by ${v^2} = 144 - 9{x^2}$. Then:
A) Displacement of the body$ < $ distance moved by body.
B) The magnitude of acceleration at a distance $3m$ from the fixed point is $27m/{s^2}$.
C) The motion is S.H.M with $T = {{2\pi } \over 3}unit$.
D) The maximum displacement from the fixed point is $4$ unit.
E) All of the above.

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Last updated date: 01st Mar 2024
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IVSAT 2024
Answer
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Hint: Solve the given equation and check for each option one by one.
Use the formula, velocity in S.H.M$v = w\sqrt {{A^2} - {x^2}} $.

Complete step by step solution:
Given: ${v^2} = 144 - 9{x^2}$
$v = \sqrt {144 - 9{x^2}}$
$\Rightarrow \sqrt {9(16 - {x^2})}$
$\Rightarrow 3\sqrt {16 - {x^2}} ....(1) $
Comparing it with the velocity in S.H.M
$v = \omega \sqrt {{A^2} - {x^2}} $
We get,
$\omega = 3$
${A^2} = 16$
$\Rightarrow A = 4m$
In S.H.M, $displacement < distance$ hence (A) is correct.
Now, $a = - \omega {x^2}$
At $x = 3, \omega = 3$
$a = - {(3)^2}(3)$
$a = - 27m/{s^2}$
$\left| a \right| = 27m/{s^2} $
Hence, (B) is also correct.
Now,
$T = {{2\pi } \over \omega}$
Putting value of $ \omega = 3$
$T = {{2\pi } \over 3}$
Hence, (C) is also correct.
The maximum displacement is amplitude = $4unit$
Hence, (D) is also correct.

Hence, all the above are correct.

Note: While solving this type question we should have a clear understanding of the entire concept so that we can apply one into the other and deduce fruitful results.