Question

Problem 4: A mass m = 1.5 kg hangs at the end of a vertical spring whose top end is fixed to the ceiling. The spring has spring constant k = 95 N/m and negligible mass. The mass undergoes simple harmonic motion when placed in vertical motion, with its position given as a function of time by y(t) = A cos(ωt – φ), with the positive y-axis pointing upward. At time t = 0 the mass is observed to be at a distance d = 0.35 m below its equilibrium height with an upward speed of v0 = 6 m/s.

Answer 1

Answer:

y (t) = 0.754 * cos ( 7.96 t - 69.52)

Explanation:

Given:

m = 1.5 kg , k = 95 N / m , v₀ = 6 m / s , d = 0.35 m , t = 0

y (t) = A * cos ( ω * t - φ )

Using the equation that describe the motion

m * v = - k * x  ⇒ m * x'' = - k * x

Angular velocity is equal to

ω = √ k / m   ⇒ ω = √ 95 N /m  /  1.5 kg  

ω = 7.96  rad /s

A = v / ω   ⇒  A = 6 m /s   /   7.96 rad / s

A = 0.754

d = cos * φ    ⇒  φ = cos ⁻¹ * 0.35

φ = 69.52

y (t) = A * cos ( ω * t - φ )    ⇒  y (t) = 0.754 * cos ( 7.96 t - 69.52)

Answer 2

Answer:

The angular frequency of spring-mass system is 7.95 rad/s.

The phase angle is 65.11 degree.

Explanation:

Given data:

The mass is, $m=1.5 \;\rm kg$.

The value of spring constant is, $k=95 \;\rm N/m$.

Distance covered at t=0 is, $y (t=0) = 0.35 \;\rm m$

Upward speed is, $v_{0}=6 \;\rm m/s$.

The position of mass is,

$y(t)=Acos(\omega t-\phi)$

Here, A is amplitude of vibration, $\omega$ is the angular frequency, t is time and $\phi$ is the phase angle.

(a)

The expression for the angular frequency of spring-mass system is:

$\omega =\sqrt\frac{k}{m}$

Substitute the values as,

$\omega =\sqrt\dfrac{95}{1.5}$

$\omega = 7.95 \;\rm rad/s$

Thus, the value of angular frequency is 7.95 rad/s.

(b)

The position of mass is, $y(t)=Acos(\omega t-\phi)$.

Differentiating the above equation to obtain the speed as,

$y(t)=Acos(\omega t-\phi)\\y'(t)= v_{0}=-A\omega sin(-\phi)$

At time t=0,

$y(t=0)=A cos(\omega (0)-\phi)\\y(t=0)=A cos(-\phi)\\45= A cos(-\phi) .....................................................(1)$

Speed of system is,

$6=-A\omega sin(-\phi) ............................................(2)$

Taking ratio of equation (2) and (1) as,

$\dfrac{-A \omega sin(- \phi)}{Acos(\phi)} = \dfrac{6}{0.35} \\\\7.95 \times tan(\phi) = \dfrac{6}{0.35} \\\phi = tan^{-1}(\dfrac{6}{0.35 \times 7.95}) \\\phi =65.11 \;\rm degree$

Thus, the phase angle is 65.11 degree.

For more details, refer to the link:

https://brainly.com/question/14670501?referrer=searchResults