# A 0.500-kg mass on a spring has velocity as a function of time given by the following equation. Find the following:

$v_x(t) = ( 2.60 cm/s) \sin \big[ ( 4.63 rad/s ) t – (\pi/2) \big]$

1. The Period
2. The Amplitude
3. Maximum Acceleration of the Mass
4. Force Constant of the Spring

The question aims to find the period, amplitude, acceleration, and force constant of the spring of a mass attached to a spring.

The question is based on the concept of simple harmonic motion (SHM). It is defined as a periodic motion of a pendulum or a mass on a spring. When it moves to and fro is called simple harmonic motion. The equation of the velocity is given as:

$v(t) = -A \omega \sin ( \omega t + \phi )$

$\omega = 4.63\ s^{-1}$

$A \omega = 2.60\ cm/s$

$\phi = \pi/2$

$m = 0.500 kg$

a) We have the value of $\omega$, so we can use its value to find the time period of the SHM. The time period T is given as:

$T = \dfrac{ 2 \pi }{ \omega }$

Substituting the values, we get:

$T = \dfrac{ 2 \pi }{ 4.63 }$

$T = 1.36\ s$

b) The given equation of the velocity above shows that the constant A before the $\sin$ represents the amplitude. Comparing the equation with the given equation of the velocity of the SHM, we get:

$A \omega = 2.60\ cm/s$

$A = \dfrac{ 2.60 \times 10^ {-2} }{ 4.63 s^{-1} }$

$A = 5.6\ mm$

c) The maximum acceleration of the mass in SHM is given by the equation as:

$a_{max} = A \times \omega^2$

Substituting the values, we get:

$a_{max} = 5.6 \times 10^{-3} \times (4.63)^2$

Simplifying the equation, we get:

$a_{max} = 0.12 m/s^2$

d) The force constant of the spring can be calculated by the given equation as:

$\omega = \sqrt{ \dfrac{ k }{ m } }$

Rearranging the equation to solve for k, we get:

$k = m \omega^2$

Substituting the values, we get:

$k = 0.500 \times (4.63)^2$

$k = 10.72\ kg/s^2$

## Numerical Result

a) Time Period:

$T = 1.36\ s$

b) The Amplitude:

$A = 5.6\ mm$

c) Maximum Acceleration:

$a_{max} = 0.12 m/s^2$

d) Force Constant of the Spring:

$k = 10.72\ kg/s^2$

## Example

A mass is attached to a spring and oscillates, making it a simple harmonic motion. The equation of the velocity is given as follows. Find the amplitude and time period of the SHM.

$v_x(t) = ( 4.22 cm/s) \sin \big[ ( 2.74 rad/s ) t – (\pi) \big]$

The value of the $\omega$ is given as:

$\omega = 2.74\ s^{-1}$

The amplitude A is given as:

$A \omega = 4.22 \times 10^{-2} m/s$

$A = \dfrac{ 4.22 \times 10^{-2} }{ 2.74 }$

$A = 15.4\ mm$

The value of the time period of the SHM is given as:

$T = \dfrac{ 2 \pi }{ \omega }$

$T = \dfrac{ 2 \pi }{ 2.74 }$

$T = 2.3\ s$