Question

The maximum speed and acceleration of a particle executing simple harmonic motion are 10 cm s⁻¹ and 50 cm s⁻². Find the position(s) of the particle when the speed is 8 cm s⁻¹.

Answer 1

It is given that:
Maximum speed of the particle, $v_{Max}$ = 10 cms^$-1$
Maximum acceleration of the particle, $a_{Max}$ = 50 cms⁻²​

The maximum velocity of a particle executing simple harmonic motion is given by,
$v_{Max}=A\omega$
where $\omega \mathrm{is}\mathrm{angular}\mathrm{frequency},\mathrm{and}$
A is amplitude of the particle.

Substituting the value of $v_{Max}$ in the above expression, we get:
Aω = 10 $...\left(1\right)$
$\Rightarrow {\mathrm{\omega }}^2=\frac{100}{A^2}$

a_Max = ω²A = 50 cms⁻¹
$$\begin{gathered} \Rightarrow {\omega }^2=\frac{50}{A}...\left(2\right) \\ \mathrm{From}\mathrm{the}\mathrm{equations}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get}: \\ \frac{100}{A^2}=\frac{50}{A} \\ \Rightarrow A=2\mathrm{cm} \\ \mathbf{\therefore }\omega =\sqrt{\frac{100}{A^2}}=5{\sec }^{-1} \end{gathered}$$

To determine the positions where the speed of the particle is 8 ms^-1, we may use the following formula:
v² = ω² (A² − y²)
where y is distance of particle from the mean position, and
v is velocity of the particle.

On substituting the given values in the above equation, we get:
64 = 25 (4 − y²)
$\Rightarrow \frac{64}{25}=4-y^2$

⇒ 4 − y² = 2.56
⇒ y² = 1.44
⇒​ y = $\sqrt{1.44}$
⇒ y = ± 1.2 cm (from the mean position)