A 0.50\text{ kg} mass attached to a horizontal spring undergoes simple harmonic motion on … — JAMB UTME

A $0.50\text{ kg}$ mass attached to a horizontal spring undergoes simple harmonic motion on a frictionless surface. The total mechanical energy of the system is $0.16\text{ J}$ and the force constant of the spring is $32\text{ N/m}$ . What is the speed of the mass, in $\text{m/s}$ , at the instant when the magnitude of its acceleration is $3.84\text{ m/s}^2$ ?

Cevap: 0.64 m/s

Cevap

The speed of the mass at that instant is

0.64\text{ m/s}

Using the relation for total mechanical energy

E = \frac{1}{2}kA^2

, the amplitude is

A = 0.10\text{ m}

. The angular frequency is

\omega = \sqrt{k/m} = 8.0\text{ rad/s}

. From

|a| = \omega^2 |x|

, the displacement magnitude when acceleration is

3.84\text{ m/s}^2

|x| = 0.06\text{ m}

. Substituting these values into

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

yields

v = 8.0 \sqrt{0.10^2 - 0.06^2} = 0.64\text{ m/s}

Adım Adım Çözüm

Calculate the angular frequency of the simple harmonic motion

\omega = 8.0\text{ rad/s}

The angular frequency depends on the stiffness constant and the mass according to \omega = \sqrt{k/m}.

Calculate the amplitude of oscillation from total energy

A = 0.10\text{ m}

The total mechanical energy in SHM is given by E = \frac{1}{2}kA^2.

Find the magnitude of displacement corresponding to the given acceleration

|x| = 0.06\text{ m}

In SHM, acceleration magnitude is related to displacement magnitude by |a| = \omega^2 |x|.

Calculate the speed at this displacement using the SHM velocity-displacement relation

v = 0.64\text{ m/s}

Velocity in SHM is calculated using v = \omega \sqrt{A^2 - x^2}.

Anahtar Kavram

Interdependence of energy, angular frequency, acceleration, and velocity in Simple Harmonic Motion

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