Physics, 28.11.2019 06:31 sydchann582

Amass 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 = 135 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 = 0.45 m below its equilibrium height with an upward speed of v0 = 5 m/s.
(a) find the angular frequency of the oscillation in radians per second. 20%
(b) find the value of, in radians.

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Answer from: jasmin5285

(a). The angular frequency is 9.48 rad/s.

(b). The value of Ф is 0.8644 radian.

Explanation:

Given that,

Mass m= 1.5 kg

Spring constant = 135 N/m

Distance = 0.45 m

Speed = 5 m/s

$y_{0}=0.45$

The position function is

$y(t)=A\cos(\omega t-\phi)$

(a). We need to calculate the angular frequency of the oscillation

Using formula of angular frequency

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

Where, k = spring constant

m = mass

Put the value into the formula

$\omega=\sqrt{\dfrac{135}{1.5}}$

$\omega=9.48\ rad/sec$

(b). We need to calculate the value of φ

Using given position function

$y(t)=A\cos(\omega t-\phi)$

$y'(t)=v(t)=-A\omega\sin(\omega t-\phi)$

At t = 0,

$y(0)=A\cos(-\phi)$

$0.45=A\cos(-\phi)$ ...(I)

At t = 0, v(0) =5 m/s

$5=-A\omega\sin(-\phi)$ ....(II)

Divide equation (II) by equation (I)

$\dfrac{5}{0.45}=9.48\tan\phi$

$\phi=\tan^{-1}(\dfrac{5}{0.45\times9.48})$

$\phi=0.8644\ radian$

Hence, (a). The angular frequency is 9.48 rad/s.

(b). The value of Ф is 0.8644 radian.

Answer from: Quest

here we know that ball will drop down by total distance 3.2 m

so here we know that

$v_{yi} = 0$

$\delta y = - 3.2 m$

now using kinematics

$\delta y = v_{yi} t + \frac{1}{2}at^2$

$-3.2 = 0 -\frac{1}{2}(9.8) t^2$

$t = 0.81 s$

now in the same time it will cover horizontal distance of 0.75 m

$\delta x = 0.75m$

so speed of the ball will be

$v = \frac{delta x }{\delta t}$

$v = \frac{0.75}{0.81} = 0.93 m/s$

Answer from: Quest

m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| = √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| = √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| = 14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s