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

**The Period****The Amplitude****Maximum Acceleration of the Mass****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:

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

## Expert Answer

The given information about this problem is as follows:

\[ \omega = 4.63\ s^{-1} \]

\[ A \omega = 2.60\ cm/s \]

\[ \phi = \pi/2 \]

\[ m = 0.500 kg \]

**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:

\[ T = \dfrac{ 2 \pi }{ \omega } \]

\[ T = \dfrac{ 2 \pi }{ 2.74 } \]

\[ T = 2.3\ s \]