An oscillator consists of a block attached to a spring (k = 299 N/m). At some time t, the position (measured from the system's equilibrium location), velocity, and acceleration of the block are x = 0.0817 m, v = -16.9 m/s, and a = -108 m/s2. Calculate (a) the frequency (in Hz) of oscillation, (b) the mass of the block, and (c) the amplitude of the motion

(a) The frequency is

f = sqrt(k/m)/(2 pi)
You know k but do not know m. Get m from
m = F/a = -k x/a = -299*0.0817/(-108)= 0.226 kg
Now you can solve for f.
(b) You have m already
(c) At max amplitude X,
(1/2) k X^2 = total energy
= (1/2)m v^2 + (1/2) k x^2
where v and x are initial values.
Solve for X

To calculate the frequency of oscillation (a), the mass of the block (b), and the amplitude (c) of the motion, we can use the following formulas:

(a) Frequency (f) = 1 / T
(b) Mass (m) = F / a
(c) Amplitude (A) = x

Let's calculate each value step by step:

(a) Frequency (f) = 1 / T
To find the frequency, we need to determine the period (T). The period is the time taken for one complete oscillation. We can calculate it using the time taken for one complete oscillation cycle, which is equal to the time period of the block.

Given:
t = ?
x = 0.0817 m
v = -16.9 m/s

Using the formula v = A * 2πf * cos(2πft), where A is the amplitude, f is the frequency, and t is the time.

Since the block is at x = 0.0817 m and moving away from equilibrium, we can assume it started from its maximum displacement. At the maximum displacement, the velocity is 0, so we can solve for t:

0 = 0.0817 * 2πf * cos(2πft)
cos(2πft) = 0

To find the time period T, we need to find the smallest positive value of t that satisfies this equation. We can solve this numerically or graphically.

(b) Mass (m) = F / a
Given:
k = 299 N/m (spring constant)
a = -108 m/s^2 (acceleration)

For a mass-spring system, the force (F) can be calculated using Hooke's Law: F = -kx, where x is the displacement from equilibrium.

On the other hand, the force can also be calculated using Newton's second law: F = ma.

Equating the two formulas, we get:
-kx = ma

Rearranging the equation, we have:
m = -kx / a = (-299 N/m * 0.0817 m) / -108 m/s^2

(c) Amplitude (A) = x
The amplitude (A) is the maximum displacement from the equilibrium position. Given in the question, the amplitude is equal to x, which is 0.0817 m.

Now let's calculate each value:

(a) Frequency (f) = 1 / T
Determine the smallest positive value of t that satisfies cos(2πft) = 0.

(b) Mass (m) = F / a
Substitute the values of k, x, and a into the equation m = -kx / a.

(c) Amplitude (A) = x
The amplitude (A) is given in the question as x = 0.0817 m.

Using these calculations, we can find the frequency (a), mass (b), and amplitude (c) of the oscillator system.