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If ' $\mathrm{v}$ ' is velocity and ' $a$ ' is acceleration of a particle executing linear simple harmonic motion. Which one of the following statements is correct?
Options:
Solution:
1563 Upvotes
Verified Answer
The correct answer is:
when ' $a$ ' is maximum, $v$ is zero
The correct option is (B).
Concept: For SHM, $\mathrm{a}=-\left(\omega^2\right) \mathrm{x}$ is the necessary condition. A general solution for SHM can be written as,
$x=A \sin (\omega t+\phi)$
where, $\mathrm{x}$ is the displacement from mean position, $A$ is the amplitude, $\omega$ is the angular frequency and $\phi$ is the phase angle.
On taking the first derivative, velocity can be written as
$$
v=\frac{d x}{d t}=(A \omega) \cos (\omega t+f)=(A w) \sin \left(\omega t+\phi+\frac{\pi}{2}\right)
$$
And, taking derivative of velocity, acceleration can be written as:
$$
\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=-\left(\mathrm{A} \omega^2\right) \sin (\omega \mathrm{t}+\phi)=\left(\mathrm{A} \omega^2\right) \sin (\omega \mathrm{t}+\phi+\pi)
$$
Therefore, in SHM acceleration is ahead of velocity by a phase $\frac{\pi}{2}$ Hence, Option (B) is the only correct option.
Concept: For SHM, $\mathrm{a}=-\left(\omega^2\right) \mathrm{x}$ is the necessary condition. A general solution for SHM can be written as,
$x=A \sin (\omega t+\phi)$
where, $\mathrm{x}$ is the displacement from mean position, $A$ is the amplitude, $\omega$ is the angular frequency and $\phi$ is the phase angle.
On taking the first derivative, velocity can be written as
$$
v=\frac{d x}{d t}=(A \omega) \cos (\omega t+f)=(A w) \sin \left(\omega t+\phi+\frac{\pi}{2}\right)
$$
And, taking derivative of velocity, acceleration can be written as:
$$
\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=-\left(\mathrm{A} \omega^2\right) \sin (\omega \mathrm{t}+\phi)=\left(\mathrm{A} \omega^2\right) \sin (\omega \mathrm{t}+\phi+\pi)
$$
Therefore, in SHM acceleration is ahead of velocity by a phase $\frac{\pi}{2}$ Hence, Option (B) is the only correct option.
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