Question

An oscillator consists of a block attached to a spring (k = 500 N/m). At some time t, the position (measured from the system's equilibrium location), velocity, and acceleration of the block are x = 0.660 m, v = -12.3 m/s, and a = -128 m/s2. (a) Calculate the frequency of oscillation. Hz (b) Calculate the mass of the block. kg (c) Calculate the amplitude of the motion.

Answer 1

Answer:

a) $\omega = 10.407\,\frac{rad}{s}$, b) $m = 4.617\,kg$, c) $A = 1.355\,m$

Explanation:

a) The system have a simple armonic motion, whose position function is:

$x(t) = A\cdot \cos (\omega\cdot t + \phi)$

The velocity function is determined by deriving the position function in terms of time:

$v(t) = -\omega \cdot A \cdot \sin(\omega\cdot t + \phi)$

The acceleration function is found by deriving again:

$a(t) = -\omega^{2} \cdot A \cdot \cos (\omega\cdot t + \phi)$

Let assume that $t = 0\,s$. The following nonlinear system is built:

$A\cdot \cos \phi = 0.660\,m$

$-\omega \cdot A \cdot \sin \phi = -12.3\,\frac{m}{s}$

$-\omega^{2}\cdot A \cdot \sin \phi = -128\,\frac{m}{s^{2}}$

System can be reduced by divinding the second and third expressions by the first expression:

$\omega \cdot \tan \phi = 18.636\,\frac{1}{s}$

$\omega^{2}\cdot \tan \phi = 193.94\,\frac{1}{s^{2}}$

Now, the last expression is divided by the first one:

$\omega = 10.407\,\frac{rad}{s}$

b) The mass of the block is:

$m = \frac{k}{\omega^{2}}$

$m = \frac{500\,\frac{N}{m} }{(10.407\,\frac{rad}{s})^{2} }$

$m = 4.617\,kg$

c) The phase angle is:

$\phi = \tan^{-1} \left(\frac{18.636\,\frac{1}{s} }{\omega} \right)$

$\phi \approx 0.338\pi$

The amplitude is:

$A = \frac{0.660\,m}{\cos 0.338\pi}$

$A = 1.355\,m$

Answer 2

Answer:

Explanation:

Given:

Spring constant, k = 500 N/m

Displacement, x = 0.660 m

Velocity, v = -12.3 m/s

Acceleration, a = -128 m/s2

For a body experiencing simple harmonic motion,

x = A cos (ωt + φ)

0.66 = A cos (ωt + φ) ....1

dx/dt = A cos (ω t + φ ) dt

dx/dt = v = -Aω × sin (ω t + φ)

-12.3 = -Aω × sin (ω t + φ) ......2

dv/dt = -Aω × sin (ω t + φ) dt

dv/dt = a = -Aω^2 × cos (ω t + φ)

-128 = -Aω^2 × cos (ω t + φ) .......3

Equating equation 1 and 3,

-128 = -ω^2 × 0.66

ω^2 = 128/0.66

= 193.94

ω = 13.93 rad/s

ω = 2pi × f

frequency, f = 13.93/2pi

= 2.22 Hz

B.

Using Hooke's law,

Force, F = -kx

Force = mass, m × acceleration, a

Mass = (500 × 0.66)/128

= 2.58 kg

C.

Amplitude, A

ω = 13.93 rad/s

Frome equation 2 and 3,

-12.3 = -Aω × sin (ω t + φ)

-12.3 = -A × 13.93 × sin (13.93 × 1/2.22 + φ)

0.883 = A × sin (6.275 + φ) .....4

-128 = -Aω^2 × cos (ω t + φ)

-128 = A (13.93)^2 cos (13.93 × 1/2.22 + φ)

0.66 = A cos (6.275 + φ) .....5

From equation 4 and 5,

0.883 = A × sin (6.275 + φ)

0.66 = A cos (6.275 + φ)

Squaring both and equating them,

0.78/A^2 = sin^2 (6.275 + φ)

0.436/A^2 = cos^2 (6.275 + φ)

Adding both,

0.78/A^2 + 0.436/A^2 = sin^2 (6.275 + φ) + cos^2 (6.275 + φ)

From sin^2 theta + cos^2 theta = 1

0.78/A^2 + 0.436/A^2 = 1

0.78 + 0.436 = A^2

A = sqrt(1.2156)

= 1.1025 m