Question

(III) A 1.60-kg object oscillates at the end of a vertically hanging light spring once every 0.45 s. (a) Write down the equation giving its position y (+upward) as a function of time t. Assume the object started by being compressed 16 cm from the equilibrium position (where y = 0), and released. (b) How long will it take to get to the equilibrium position for the first time? (c) What will be its maximum speed? (d) What will be the object’s maximum acceleration, and where will it first be attained?

Short answer

(i) The equation giving the position of the object as a function of time t is:

\(y = \left( {0.16\;{\rm{m}}} \right)\cos \left( {14t} \right)\).

(ii) It will take 0.11 s for the object to get to the equilibrium position for the first time.

(iii) The maximum speed of the object will be \({\rm{2}}{\rm{.2}}\;{\rm{m/s}}\).

(iv)The maximum acceleration of the object will be 31 m/s² and it will be first attained at the release point.

Step-by-step solution

Simple harmonic motion as sinusoidal motion

Simple harmonic motion is expressed as sinusoidal motion because the position of an object varies as a sinusoidal function of time.

When an oscillating object starts moving from rest from the point of its maximum displacement towards the equilibrium position, then the displacement of the object is defined by the cosine function.

Identification of the given information

The mass of the object is, m = 1.60 kg.

The time period of the oscillation of the spring is, T = 0.45 s.

The amplitude of the object is, \(A = 16\;{\rm{cm}}\).

Determination of the equation of position as a function of time

Since the displacement of spring starts from the position of its maximum displacement in the positive direction, therefore position y at any time t will be represented by the cosine function as:

\(y = A\cos (\omega t)\) ….. (i)

Here,\(\omega \)is the angular velocity.

The angular velocity of the object is given as:

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

On substituting the value of T, you will get:

\(\begin{aligned}{c}\omega &= \frac{{2 \times 3.14}}{{0.45\;{\rm{s}}}}\\ &= 13.96\;{\rm{rad/s}}\\ \approx {\rm{14}}\;{\rm{rad/s}}\end{aligned}\)

Now, on putting the values of A and \(\omega \) in equation (i), you will get:

\(\begin{aligned}{l}y &= \left( {16\;{\rm{cm}} \times \frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)\cos \left( {14t} \right)\\y &= \left( {0.16} \right)\cos \left( {14t} \right)\end{aligned}\)

This is the required equation of position of the object as a function of time.

Determination of time to reach the equilibrium position for the first time

The time taken (t) by the object to reach the equilibrium position for the first time will be one fourth of the time period (T), that is,

\(\begin{aligned}{c}t &= \frac{T}{4}\\ &= \frac{{0.45\;{\rm{s}}}}{4}\\ &= 0.11\;{\rm{s}}\end{aligned}\)

Thus, the time taken by the object to reach at its equilibrium position is 0.11 s.

Determination of the maximum speed

The maximum speed \(\left( {{v_{\max }}} \right)\) of an object oscillating with frequency f in simple harmonic motion is given as:

\(\begin{aligned}{c}{v_{\max }} &= 2\pi fA\\ &= \frac{{2\pi }}{T}A\\ &= \omega A\end{aligned}\)

On putting the given values, you will get:

\(\begin{aligned}{c}{v_{\max }} &= (13.96\;{\rm{rad}}/{\rm{s}})(0.16\;{\rm{m}})\\ &= 2 \cdot 234\;{\rm{m}}/{\rm{s}}\\ \approx {\rm{2}}{\rm{.2}}\;{\rm{m/s}}\end{aligned}\)

Thus, the maximum speed of the object will be \({\rm{2}}{\rm{.2}}\;{\rm{m/s}}\).

Determination of the object’s maximum acceleration

The maximum acceleration \(\left( {{a_{\max }}} \right)\) of an object in simple harmonic motion is given as:

\({a_{{\rm{max }}}} = {\omega ^2}A\)

On substituting the given values, you will get:

\(\begin{aligned}{c}{a_{\max }} &= {(13.96\;{\rm{rad/s}})^2}(0.16\;{\rm{m}})\\ &= 31.18\;{\rm{m/}}{{\rm{s}}^2}\\ \approx 31\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\end{aligned}\)

Thus, the maximum acceleration of the object will be \(31\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\). It is attained at the extreme points of the oscillations. Since the object is released from the extreme point, therefore the maximum acceleration will be first attained at the release point itself.