A particle moves with simple harmonic motion in a straight line. In first $t s$ , after starting from rest it travels a distance $a$ and in next $t s$ , it travels $2 a$ in same direction. The time period of the oscillation is found to be $n t$ , then $n$ is

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Solution

Consider a general equation, $x = A c o s ω t$
Apply for OP, $A - a = A c o s ω t$
From here, $c o s ω t = \frac{A - a}{A}$

Similarly, $A - (a + 2 a) = A c o s 2 ω t$
[ $∴$ time doubled]
$c o s 2 ω t = \frac{A - 3 a}{A} c o s 2 θ = 2 {c o s}^{2} θ - 1 {(\frac{A - a}{A})}^{2} = \frac{A - 3 a + A}{2 A}$

From here, $A = 2 a$
So, amplitude of the motion is $2 a$

$2 a - a = 2 a c o s ω t \frac{π}{3} = ω t \frac{π}{3} = \frac{2 π}{T} t T = 6 t$
$⟹ n = 6$

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A particle moves with simple harmonic motion in a straight line. In first t s, after starting from rest it travels a distance a and in next t s, it travels 2 a in same direction. The time period of the oscillation is found to be nt, then n is

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A particle moves with simple harmonic motion in a straight line. In first $t s$ , after starting from rest it travels a distance $a$ and in next $t s$ , it travels $2 a$ in same direction. The time period of the oscillation is found to be $n t$ , then $n$ is