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A massless spring hangs from the ceiling with a small object attached to its lower end. The object is initially held at rest in a position yi such that the spring is at its rest length. The object is then released from yi and oscillates up and down, with its lowest position being 10 cm below yi.
(a) What is the frequency of the oscillation?
(b) What is the speed of the object when it is 8.0 cm below the initial position?
(c) An object of mass 300 g is attached to the first object, after which the system oscillates with half the original frequency. What is the mass of the first object?
(d) How far below yi is the new equilibrium (rest) position with both objects attached to the spring?

Answer :

usmanbiu

Answer:

(A) 1.58 Hz

(B) 0.99 m/s

(C) 0.1 kg

(D) 0.4 m

Explanation:

extension of the spring (x) = 10 cm = 0.1 m

acceleration due to gravity (g) = 9.8 m/s^{2}

(A) force = mg = kx

     where m = mass

     g = acceleration due to gravity

     x = extension

    mg = kx  

    substituting the values 0f g and x we have

     9.8m = 0.1k

    therefore k = 9.8m/0.1

    k = 98m

    formula for frequency (f) = [tex]\frac{1}{2π}.\sqrt{\frac{k}{m} }[/tex]

    inserting the value of spring constant (k) as 98m into the equation above

    f = [tex]\frac{1}{2π}.\sqrt{\frac{98m}{m} }[/tex]

    f = [tex]\frac{1}{2π}.\sqrt{98}[/tex]

    f = 1.58 Hz

(B) find the speed of the object when it is 8 cm below its initial position

   

    from the conservation of energy,

   initial potential energy (U) + kinetic energy (K.E) = 0

    ([tex]0.5ky^{2} - mgy) + 0.5mv^{2}[/tex]

     v = [tex]\sqrt{2gy - \frac{k}{m}.y^{2}}[/tex]

     where y = position of the spring = 8 cm (0.08m) and k = 98m as in (A) above

     v =  [tex]\sqrt{2 x 9.8 x 0.08 - \frac{98m}{m}.0.08^{2}}[/tex]

     v =  [tex]\sqrt{1.568 - (98 x 0.0064}[/tex]

     v = 0.99 m/s

(C) find the mass (m) of the object when an object of mass 300 g is attached to the first object, after which the system oscillates with half the original frequency.

after addition of the 300 g mass

new frequency = half the initial frequency

[tex]\frac{1}{2π}.\sqrt{\frac{k}{m + 300} }[/tex] = 0.5 x  [tex]\frac{1}{2π}.\sqrt{\frac{k}{m} }[/tex]

[tex]\sqrt{\frac{k}{m + 300} }[/tex] = 0.5 x  [tex]\sqrt{\frac{k}{m} }[/tex]

[tex](\sqrt{\frac{k}{m + 300} })^{2} =(0.5 x \sqrt{\frac{k}{m} })^{2}[/tex]

[tex]\frac{k}{m + 300} =0.25 x \frac{k}{m}[/tex]

[tex]\frac{1}{m + 300} =0.25 x \frac{1}{m}[/tex]

0.25 (m + 300) = m

m - 0.25m = 0.25 x 300

m- 0.25m = 75

0.75 m = 75

m = 100 g = 0.1 kg  

(D) find the new equilibrium position

   from mg = kx we can find the new equilibrium position (x)

   where m = m + 300 = m + 0.3 (in kg)

     (m+0.3)g = kx

   x = [tex]\frac{(m+0.3)g}{k}[/tex]

  recall that k = 98m

    x =  [tex]\frac{(m+0.3)g}{98m}[/tex]

    now substituting the values of m and g into the equation

  x =  [tex]\frac{(0.1 + 0.3) x 9.8}{98 x 0.1}[/tex]

  x = 0.4 m

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