Please. Answer parts please. Attempt Once The maximum velocity attained by the mass of a simple harmonic oscillator is10 cm/sec, and the period of oscillation is 2 sec. If the mass is released withan initial displacement of 2 cm, find (a) the amplitude, (b) the initial velocity,(c) the maximum acceleration, and (d) the phase angle.

Answered: The maximum velocity attained by the…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Please. Answer parts please. Attempt Once

The maximum velocity attained by the mass of a simple harmonic oscillator is
10 cm/sec, and the period of oscillation is 2 sec. If the mass is released with
an initial displacement of 2 cm, find (a) the amplitude, (b) the initial velocity,
(c) the maximum acceleration, and (d) the phase angle.

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(a)

To find:

The amplitude of the oscillation:

Given :

The velocity is $v = 10 \frac{cm}{s}$ .

The period of oscillation is $t = 2 \sec$ .

The initial displacement is $x = 2 cm$ .

Formula used:

The angular velocity of oscillation is,

$ω = \frac{2 π}{t}$

Here, $ω$ is angular velocity.

The amplitude $A$ of the the oscillation is

$v = A ω$

Here, $v$ is velocity.

Calculation:

The angular velocity of oscillation can be obtained as,

$ω = \frac{2 π}{2} = π rad / s$

The amplitude $A$ of the the oscillation can be obtained as,

$\begin{array}{rcl} 10 \frac{cm}{s} (\frac{1 m}{100 cm}) & = & \frac{A}{π} \\ A & = & 0.03183 m \end{array}$

Thus, the amplitude $A$ of the the oscillation is $\begin{array}{rcl} 0.03183 m \end{array}$ .

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