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

Step-by-step 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

Answer 2

a)
`\omega = 10.407\,(rad)/(s)`, b)
`m = 4.617\,kg`, c)
`A = 1.355\,m`

Step-by-step 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\,(m)/(s)`

`-\omega^(2)\cdot A \cdot \sin \phi = -128\,(m)/(s^(2))`

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

`\omega \cdot \tan \phi = 18.636\,(1)/(s)`

`\omega^(2)\cdot \tan \phi = 193.94\,(1)/(s^(2))`

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

`\omega = 10.407\,(rad)/(s)`

b) The mass of the block is:

`m = (k)/(\omega^(2))`

`m = (500\,(N)/(m) )/((10.407\,(rad)/(s))^(2) )`

`m = 4.617\,kg`

c) The phase angle is:

`\phi = \tan^(-1) \left((18.636\,(1)/(s) )/(\omega) \right)`

`\phi \approx 0.338\pi`

The amplitude is:

`A = (0.660\,m)/(\cos 0.338\pi)`

`A = 1.355\,m`