Answer: 1.8 s
Step-by-step explanation:
Given
The height of the spring mass system above the table is given by
[tex]h(t)=6.85\cos (3.42t)+9[/tex]
The mass is performing S.H.M with frequency [tex]\omega =3.42[/tex]
and [tex]\omega T=2\pi[/tex]
[tex]\therefore T=\dfrac{2\pi }{\omega}[/tex]
time when mass returns to its original position
[tex]T=\dfrac{2\pi }{3.42}\\\\T=1.8\ s[/tex]
The number of seconds, to the nearest tenth, will it take for the mass to return to its original position is 1.8 seconds.
Since the equation h(t)=6.85cos(3.42t)+9, where the height is in inches and the time, t, is in seconds
So here the time should be
[tex]= 2\pi \div 3.42\\\\= 2\times 3.14\div 3.42[/tex]
= 1.8 seconds
Therefore, The number of seconds, to the nearest tenth, will it take for the mass to return to its original position is 1.8 seconds.
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