An object of mass 0.20\text{ kg} executes simple harmonic motion with an amplitude of 0.05… — JAMB UTME

An object of mass $0.20\text{ kg}$ executes simple harmonic motion with an amplitude of $0.05\text{ m}$ and a maximum acceleration of $20\text{ m/s}^2$ . What is the speed of the object when its displacement from the equilibrium position is $0.03\text{ m}$ ?

A
$0.16\text{ m/s}$
B
$0.60\text{ m/s}$
$0.80\text{ m/s}$ Cevap
D
$1.00\text{ m/s}$

Cevap

The speed of the object at a displacement of

0.03\text{ m}

0.80\text{ m/s}

The maximum acceleration in simple harmonic motion is given by

a_{\text{max}} = \omega^2 A

. Substituting

a_{\text{max}} = 20\text{ m/s}^2

and

A = 0.05\text{ m}

gives

\omega^2 = 400\text{ rad}^2/\text{s}^2

, so

\omega = 20\text{ rad/s}

. The speed at any displacement

x

is given by

v = \omega \sqrt{A^2 - x^2}

. For

x = 0.03\text{ m}

v = 20 \sqrt{0.05^2 - 0.03^2} = 20 \times 0.04 = 0.80\text{ m/s}

Adım Adım Çözüm

Determine the angular frequency (

\omega

) of the simple harmonic motion from the maximum acceleration formula.

\omega = 20\text{ rad/s}

Maximum acceleration is given by

a_{\text{max}} = \omega^2 A

. Rearranging gives

\omega^2 = \frac{a_{\text{max}}}{A} = \frac{20}{0.05} = 400\text{ rad}^2/\text{s}^2

, so

\omega = 20\text{ rad/s}

Calculate the speed (

v

) at the given displacement (

x = 0.03\text{ m}

) using the SHM velocity formula.

v = 0.80\text{ m/s}

The speed at displacement

x

v = \omega \sqrt{A^2 - x^2} = 20 \times \sqrt{0.05^2 - 0.03^2} = 20 \times \sqrt{0.0016} = 20 \times 0.04 = 0.80\text{ m/s}

Anahtar Kavram

Simple Harmonic Motion Velocity and Acceleration Relationships

Tahmini Süre:1m 30s

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