The acceleration displacement graph of a particle executing simple harmonic motion is shown in fig. The time period of simple harmonic motion is

Question

The acceleration displacement graph of a particle executing simple harmonic motion is shown in fig. The time period of simple harmonic motion is <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/91a7ccb9a7dd63f67fe3a3847c2fcf76-.png" /> (A) (4 π/√3) s (B) (2 π/√3) s (C) The given graph doesn't represent simple harmonic motion. (D) Information is insufficient.

Tardigrade Admin · Accepted Answer

(d a/d x)=- tan 37°=-(3/4) ⇒ a=-(3/4) x so, ω2=(3/2) ⇒ (2 π/n)=(√3/6) ⇒ T=(4 π/5) s

Q. The acceleration displacement graph of a particle executing simple harmonic motion is shown in fig. The time period of simple harmonic motion is

$\frac{4 π}{3} s$

$\frac{2 π}{3} s$

The given graph doesn't represent simple harmonic motion.

Information is insufficient.

$\frac{d a}{d x} = - tan 3 7^{\circ} = - \frac{3}{4} \Rightarrow a = - \frac{3}{4} x$
so, $ω^{2} = \frac{3}{2} \Rightarrow \frac{2 π}{n} = \frac{3}{6} \Rightarrow T = \frac{4 π}{5} s$

Q. The acceleration displacement graph of a particle executing simple harmonic motion is shown in fig. The time period of simple harmonic motion is

3​4π​s

3​2π​s

The given graph doesn't represent simple harmonic motion.

Information is insufficient.

dxda​=−tan37∘=−43​⇒a=−43​x so, ω2=23​⇒n2π​=63​​⇒T=54π​s

$\frac{4 π}{3} s$

$\frac{2 π}{3} s$

$\frac{d a}{d x} = - tan 3 7^{\circ} = - \frac{3}{4} \Rightarrow a = - \frac{3}{4} x$
so, $ω^{2} = \frac{3}{2} \Rightarrow \frac{2 π}{n} = \frac{3}{6} \Rightarrow T = \frac{4 π}{5} s$