Question

An object attached to a spring vibrates with simple harmonic motion as described by the figure below. 

(a) For this motion, find the amplitude.   

(b) For this motion, find the period. 

(c) For this motion, find the angular frequency.  

(d) For this motion, find the maximum speed  

(e) For this motion, find the maximum acceleration.  

(f) For this motion, find an equation for its position x in terms of a sine function. 

Answer 1

Concepts and reason

The concept used to solve the problem is spring vibrating in a simple harmonic motion.

Initially, the amplitude and time period of the spring in Simple Harmonic Motion is determined from the figure provided. Later, angular frequency and maximum speed of the spring constant is calculated. Then, the maximum acceleration of the spring constant is calculated. Finally, the equation for the oscillating spring is determined.

Fundamentals

Amplitude of an object is defined as the maximum displacement made by the object. It is denoted as A.

Time period is the time taken by the object to complete one cycle or one oscillation. It is denoted as T.

The angular frequency of the spring in Simple Harmonic Motion is given as:

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

Here, T is the time period of the object.

The maximum speed of the object is expressed as:

$v = A\omega$

Here, $A$ is the amplitude of the object and $\omega$ is the angular frequency of the object.

The maximum acceleration of the object is expressed as:

$a = A{\omega ^2}$

Here, $A$ is the amplitude of the object and $\omega$ is the angular frequency of the object.

The equation of motion for the object in a Simple Harmonic motion in a Sine function as:

$x = A{\mathop{\rm Sin}\nolimits} \left( {\omega t + \phi } \right)$

Here, A is the amplitude, $\omega$ is the angular frequency and $\phi$ is the phase of the wave.

(a)

The graph provided is a Displacement-Time Graph. The x-axis provides the displacement made by the object and the y-axis gives the time taken by the object to travel.

Amplitude of an object is defined as the maximum displacement made by the object. From the graph provided, the maximum displacement made by the object is 2cm.

Hence, as per the definition of Amplitude, the Amplitude of the object is 2cm.

(b)

The object in a Simple Harmonic Motion moves from mean position to extreme positions. As the object moves from mean position to extreme position and reaches back to mean position, one oscillation or one cycle is said to be completed.

The graph represents the positive displacement when it moves to extreme position and the negative displacement as it returns from the extreme position.

The graph provided shows that for a complete one cycle or oscillation, the time taken is 4seconds.

Hence, as per the definition of Time period, the time period is 4 seconds. Therefore, the time period as:

$T = 4\;\sec$

(c)

The angular frequency of the object is given as:

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

Substitute 4 sec for T and find the angular frequency.

$\begin{array}{c}\\\omega = \frac{{2\pi }}{T}\\\\ = \frac{{2\pi }}{{4\;\sec }}\\\\ = 1.57\;{\rm{rad/sec}}\\\end{array}$

Therefore, the angular frequency of the object is $1.57\;{\rm{rad/sec}}$ .

(d)

Maximum speed of the object is given as:

$v = A\omega$

Substitute $1.57\;{\rm{rad/sec}}$ for $\omega$ , 2 cm for Amplitude of the object, to find the maximum speed.

$\begin{array}{c}\\v = A\omega \\\\ = \left( {2\;{\rm{cm}}} \right)\left( {1.57\;{\rm{rad/sec}}} \right)\\\\ = \left( {0.02\;{\rm{m}}} \right)\left( {1.57\;{\rm{rad/sec}}} \right)\\\\ = 0.0314\;{\rm{m/sec}}\\\end{array}$

The maximum speed of the object is $0.0314\;{\rm{m/sec}}$

(e)

Maximum acceleration of the object is given as:

$a = A{\omega ^2}$

Substitute $1.57\;{\rm{rad/sec}}$ for $\omega$ , 2 cm for Amplitude of the object, to find the maximum speed.

$\begin{array}{c}\\a = A{\omega ^2}\\\\ = \left( {2\;{\rm{cm}}} \right){\left( {1.57\;{\rm{rad/sec}}} \right)^2}\\\\ = \left( {0.02\;{\rm{m}}} \right){\left( {1.57\;{\rm{rad/sec}}} \right)^2}\\\\ = 0.04929\;{\rm{m/se}}{{\rm{c}}^2}\\\end{array}$

The maximum acceleration of the object is $0.04929\;{\rm{m/se}}{{\rm{c}}^2}$ .

(f)

The equation of motion for the object in a Simple Harmonic motion in a Sine function as:

$x = A{\mathop{\rm Sin}\nolimits} \left( {\omega t + \phi } \right)$

Substitute 2cm for A, 0 for $\phi$ , $1.57\;{\rm{rad/sec}}$ for $\omega$ and find the equation for Simple Harmonic Motion.

$\begin{array}{c}\\x = A{\mathop{\rm Sin}\nolimits} \left( {\omega t + \phi } \right)\\\\ = \left( {2\;{\rm{cm}}} \right){\mathop{\rm Sin}\nolimits} \left( {1.57t + 0} \right)\\\\ = \left( {0.02\;{\rm{m}}} \right){\mathop{\rm Sin}\nolimits} \left( {1.57t} \right)\\\end{array}$

Ans: Part a

Amplitude of the object is 2 cm.

Part b

Time period of the object is 4 seconds.

Part c

The angular frequency of the object is $1.57\;{\rm{rad/sec}}$ .

Part d

The maximum speed of the object is $0.0314\;{\rm{m/sec}}$ .

Part e

The maximum acceleration of the object is $0.04929\;{\rm{m/se}}{{\rm{c}}^2}$ .

Part f

The equation of motion of the object is $\left( {0.02\;{\rm{m}}} \right){\mathop{\rm Sin}\nolimits} \left( {1.57t} \right)$ .