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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 = 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.

Respuesta :

Answer:

(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

[tex]y_{0}=0.45[/tex]

The position function is

[tex]y(t)=A\cos(\omega t-\phi)[/tex]

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

Using formula of angular frequency

[tex]\omega=\sqrt{\dfrac{k}{m}}[/tex]

Where, k = spring constant

m = mass

Put the value into the formula

[tex]\omega=\sqrt{\dfrac{135}{1.5}}[/tex]

[tex]\omega=9.48\ rad/sec[/tex]

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

Using given position function

[tex]y(t)=A\cos(\omega t-\phi)[/tex]

[tex]y'(t)=v(t)=-A\omega\sin(\omega t-\phi)[/tex]

At t = 0,

[tex]y(0)=A\cos(-\phi)[/tex]

[tex]0.45=A\cos(-\phi)[/tex]...(I)

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

[tex]5=-A\omega\sin(-\phi)[/tex]....(II)

Divide equation (II) by equation (I)

[tex] \dfrac{5}{0.45}=9.48\tan\phi[/tex]

[tex]\phi=\tan^{-1}(\dfrac{5}{0.45\times9.48})[/tex]

[tex]\phi=0.8644\ radian[/tex]

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

(b). The value of Ф is 0.8644 radian.