Question

A buoy floating in the sea is bobbing in simple harmonic motion with period 2 seconds and amplitude 8 in. Its displacement d from sea level at time t=0 seconds is 0 in, and initially it moves downward. (Note that downward is the negative direction.) Give the equation modeling the displacement d as a function of time t. d= ク

Answer 1

Answer:

The equation of the displacement $d$ as a function of time $t$ is :

$d(t)=8sin(\pi t+\pi )$

Step-by-step explanation:

Generally , A simple harmonic wave is a sinusoidal function that is it can be expressed in simple $sin$ or $cos$ terms.

Thus,

$d(t) = Asin(wt+c)$

is the general form of displacement of a SHM.

where,

• $d(t)$ is the displacement with respect to the mean position at any time $t$
• $A$ is amplitude
• $w$ is the natural frequency of oscillation ($rads^{-1}$)
• $c$ is the phase angle which indicates the initial position of the object in SHM ($rad$)

given,

1. Time period ($T$) = $2s$
2. $A=8$
3. The natural frequency ($w$) and time period ($T$) is :

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

∴

$w$ = $\frac{2\pi }{2} = \pi$ $rads^{-1}$

∴

the equation :

⇒$d(t)=8sin(\pi t+c)$                        ------1

since $d=0$ when $t=o$ ,

⇒$0=8sinc\\c=n\pi$                         ------2

where n is an integer ;

⇒since the bouy immediately moves in the negative direction , $x$ must be negative or c must be an odd multiple of $\pi$.

⇒ $d(t) = 8sin(\pi t+(2k+1)\pi )$         ------3

where k is also an integer ;

the least value of $k=0$;

thus ,

the equation is :

$d(t)=8sin(\pi t+\pi )$