Horizontal Spring - Block Oscillator

Another type of simple harmonic motion system is the horizontal spring-block oscillator. In this set-up, the spring is held in horizontal direction, thus, the object moving in simple harmonic moves from right to left or east to west repeatedly. A spring in equilibrium position is attached to a wall, and an object with a mass, \(m\) applies a force stretching the spring to a maximum displacement, \(A\). Oppositely, a restoring force from the spring pulls the object back to equilibrium position. This horizontal motion then repeats for a period of time and thus executing simple harmonic motion.

The simple harmonic motion horizontal spring-block oscillation depends on the mass of the object and the force of the spring. It shows the relationship between the angular frequency, frequency, and the period of oscillation. The angular frequency is defined by the mass of the object and the spring constant. Mathematically, it can be expressed as

\(\omega=\sqrt{\frac{k}{m}}\).

Using the relationship between the period and angular frequency, the period can be written as

\(T=\frac{2\pi}{\omega}\)

substituting \(\omega=\sqrt{\frac{k}{m}}\), the above equation can be expressed as

\(T=2\pi \sqrt{\frac{m}{k}}\).

The total energy of a horizontal-spring block oscillator is constant. However, the continuing motion indicates the transfer of kinetic energy to elastic potential energy, vice versa, repeatedly. Thus, the total energy is given by

\(E_T=KE=EPE\)

substituting \(KE=\frac 12 mv^2\) and \(EPE=\frac 12 kx^2\)

\(E_T=\frac 12 mv^2=\frac 12 kx^2\).


Example 1.

A spring is attached horizontally to a wall. When a \(6.0-kg\) object is attached at the end of the spring, what is the period of oscillation if the spring is stretched to a maximum displacement of \(0.5\; m\). Calculate for the angular frequency given that the spring constant is \(500 \;N/m\).

Given:

\(m=6.0\;kg\\ x=0.5\;m\\ k=500\;N/m\)


Solution 1.

Solve for the period using \(T=2\pi \sqrt{\frac{m}{k}}\).

\(T=2\pi \sqrt{\frac{6.0\;kg}{500\;N/m}}=0.69\;s\)

hence, it takes 0.69 s to complete one cycle of back and forth motion.


Solution 2.

 The angular frequency can be solved by the equation \(T=\frac{2\pi}{\omega}\).

\(\omega=\frac{2\pi}{T}=\frac{2\pi}{0.69\;s}=9.1\;rad/s\).

 

Or it can be done using the formula for angular frequency \(\omega=\sqrt{\frac{k}{m}}\).

\(\omega=\sqrt{\frac{500N/m}{6.0\;kg}}=9.1\;rad/s\).


Example 2.

A horizontal spring with a 10-kg object at its end is stretched to 2.0 m by a force of 80 N. a. Find the spring constant. b. What is the total energy of the system. c. How much time does it take for one complete oscillation cycle?

Given:

\(m=10\;kg\\ x=2.0\;m\\ F=80\;N\)


Solution 1.

The spring constant can be solved using the Hooke's Law \(F=kx\).

\(k=\frac {F}{x}=\frac {80\;N}{2.0\;m}=40\;N/m\).

Hence, the spring constant is equal to \(40\;N/m\). 


Solution 2.

To solve for the total energy of the system, we could use the formula for the elastic potential energy \(EPE=\frac 12kx^2\).

\(E_T=EPE=\frac 12(40\;N/m)(2.0\;m)^2=80\;J\).

Thus, the amount of energy conserved all throughout the simple harmonic motion is \(80\;J.\)


Solution 3.

The time it takes to complete one cycle of oscillation is the period of the simple harmonic motion given by \(T=2\pi \sqrt{\frac{m}{k}}\).

\(T=2\pi \sqrt{\frac{10\;kg}{40\;N/m}}=3.14\;s\).

Therefore, it takes \(3.14 \;s\) to complete one oscillation cycle.