A 0.500-kg mass on a spring has velocity as a

Danelle Albright

Danelle Albright

Answered question

2022-01-05

A 0.500-kg mass on a spring has velocity as a function of time given by vx(t)=(3.60cms})sin[(4.71rads)t(π2)]. What are (a) the period; (b) the amplitude; (c) the maximum acceleration of the mass; (d) the force constant of the spring?

Answer & Explanation

Ben Owens

Ben Owens

Beginner2022-01-06Added 27 answers

Step 1
Given that the mass of the object is m=0.5kg.
The velocity of the object as a function of time given by
vx(t)=(3.6cms1)sin[(4.71rads1)tπ2] ​(1)
Since we know that the displacement in SHM is given by
x=Acos[ωtϕ] (2)
By differentiating above equation with respect to time we get the velocity of the object in SHM, so
vx(t)=Aωsin[ωtϕ] (3)
Now by comparing the equation (1) and (3), we get
Aω=3.6cms1 (4)
ω=4.7rads1 (5)
and
ϕ=π2rad (6)
(a) The period of SHM is given by
P=2πradω
=2πrad4.71rads1
=1.33s (7)
Matthew Rodriguez

Matthew Rodriguez

Beginner2022-01-07Added 32 answers

Step 2
A=3.6cms1ω
=3.6cms14.71rad1
=0.76cm (8)

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