# P 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. - Physics

Sum

The maximum speed and acceleration of a particle executing simple harmonic motion are 10 cm/s and 50 cm/s2. Find the position(s) of the particle when the speed is 8 cm/s.

#### Solution

It is given that:
Maximum speed of the particle, $v_{Max}$= 10 "cm"^(- 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 $\text { omega is angular frequency, and }$
is amplitude of the particle.
Substituting the value of $v_{Max}$in the above expression,
we get :
= 10     $. . . (1)$
$\Rightarrow \omega^2 = \frac{100}{A^2}$

aMax = ω2A = 50 cms−1

$\Rightarrow \omega^2 = \frac{50}{A} . . . (2)$

$\text { From the equations (1) and (2), we get:}$

$\frac{100}{A^2} = \frac{50}{A}$

$\Rightarrow A = 2 cm$

$\therefore \omega = \sqrt{\frac{100}{A^2}} = 5 \sec^{- 1}$

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)

Is there an error in this question or solution?

#### APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 12 Simple Harmonics Motion
Q 4 | Page 252
Share