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A partridge of mass 4.90 kg is suspended from a pear tree by an ideal spring of negligible mass. When the partridge is pulled down 0.100 m below its equilibrium position and released, it vibrates with a period of 4.19 s .

(a) What is its speed as it passes through the equilibrium position?
(b) What is its acceleration when it is 0.050 m above the equilibrium position?
(c) When it is moving upward, how much time is required for it to move from a point 0.050 m below its equilibrium position to a point 0.050 m above it?
(d)The motion of the partridge is stopped, and then it is removed from the spring. How much does the spring shorten?

Respuesta :

Answer:

Part a)

[tex]v = 0.15 m/s[/tex]

Part b)

[tex]a = 0.1125 m/s^2[/tex]

Part c)

[tex]t = 0.7 s[/tex]

Part d)

[tex]x = 4.37 m[/tex]

Explanation:

Time period of the spring is given as

[tex]T = 2\pi\sqrt{\frac{m}{k}}[/tex]

now we know that

[tex]T = 4.19 s[/tex]

m = 4.90 kg

so we have now

[tex]4.19 = 2\pi\sqrt{\frac{4.90}{k}}[/tex]

[tex]k = 11 N/m[/tex]

Part a)

Amplitude = 0.100

[tex]\omega = \frac{2\pi}{T}[/tex]

[tex]\omega = \frac{2\pi}{4.19}[/tex]

[tex]\omega = 1.5 rad/s[/tex]

So speed at mean position is given as

[tex]v = A\omega[/tex]

[tex]v = 0.100(1.5) = 0.15 m/s[/tex]

Part b)

Force on it when it will reach the 0.050 m from equilibrium position is given as

[tex]F = kx[/tex]

[tex]a = \omega^2 x[/tex]

[tex]a = (1.5)^2(0.05)[/tex]

[tex]a = 0.1125 m/s^2[/tex]

Part c)

As we know that object is performing SHM so here we have

[tex]x = A sin\omega t[/tex]

[tex]0.05 = 0.100 sin\omega t[/tex]

[tex]\omega t = \frac{\pi}{6}[/tex]

[tex]t = 0.35 s[/tex]

so total time taken by object to move from 0.05 m below to 0.05 m above the mean position is 0.35 + 0.35 = 0.7 s

Part d)

As we know by force equation

mg = kx

[tex]4.90(9.81) = 11 x[/tex]

[tex]x = 4.37 m[/tex]

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