The angular frequency of this motion is 5.46 rad/s.
The oscillation of spring is an example of Simple Harmonic Motion(SHM).
The general equation of an SHM is given by the formula.
X = Acos(wt)
Here A is the amplitude
ω is the angular frequency
T is the time
Comparing the above equation with the given condition,
X = 17.4 cm cos(5.46t)
A = 17.4 cm
ω = 5.46 rad/s
T = 1 s
Hence, the angular frequency of this motion is 5.46 rad/s.
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For this case we have that by definition, the momentum equation is given by:

Where:
m: It is the mass
v: It is the velocity
According to the data we have:

Substituting:

On the other hand, if we clear the variable "mass" we have:

According to the data we have:

Thus, the mass is 
Answer:

Answer: hello your question is incomplete attached below is the complete question
answer : 1/2 KD^2 ( option A )
Explanation:
P.E ( potential energy ) = mgd
In case 1 P.E = 0 i.e. mgd = 0
Given that in case 2 the Mass M had moved through the Distance D by the compression of the spring
<u>The potential energy of the M in case 2 </u>
= P.E of M at rest + P.E of the spring
= 0 + 1/2 KD^2
Answer:
v(t) = s′(t) = −9sin(t)+9cos(t)
a(t) = v′(t) = −9cos(t) −9sin(t)
Explanation:
Given that
s = 9 cos(t) + 9 sin(t), t ≥ 0
Then acceleration and velocity is
v(t) = s′(t) = −9sin(t)+9cos(t)
a(t) = v′(t) = −9cos(t) −9sin(t)
Energy and matter can neither be created or destroyed.