6. A turning fork oscillates at a frequency of 441 Hz and the tip of each prong

Question

6. A turning fork oscillates at a frequency of 441 Hz and the tip of each prong moves mto either side of center. Calculate (a) the maximum speed and where it occurs (b) where it occurs when it is the half maximum speed. 7. A solid cylinder of radius 0.55 m is released from rest from a height of 3.5 m and rolls down the incline as shown. What is the angular speed of the cylinder when it reaches the horizontal surface? I = ½ MR2 A block of mass M is hung by ropes as shown. The system is in equilibrium. The point O represents the knot, the junction of the three ropes. Which of the following statements is true concerning the magnitudes of the three forces in equilibrium? (show the calculations) 8. ? . 30°

Explanation / Answer

6. The angular frequency of oscillation, w = 2?f

Thus, w = (2×3.14×441) rad/s = 2769.48 rad/s

Amplitude of oscillation, A = 1.5 mm = 1.5×10^-3 m

Now, speed, v = w?(A^2 - x^2), where x is the displacement from equillibrium position.

Maximum speed will occur at the equillibrium position where x = 0 and is given by,

Vm = wA = (2769.48×1.5×10^-3) m/s = 4.15 m/s

(b) Let at a displacemnt x from equillibrium position the speed become half the maximum value, that is, speed will be = Vm/2 = 4.15/2 = 2.075 m/s

Thus we have, 2.075 = w?(A^2 - x^2)

   => (2.075)^2 = w^2 (A^2 - x^2)

=> 4.3056 = w^2 (A^2 - x^2)

   => 4.3056 = (2769.48)^2 [(1.5×10^-3)^2 - x^2]

=> 4.3056/7670019.47 = 2.25×10^-6 - x^2

=> 0.56×10^-6 = 2.25×10^-6 - x^2

=> x^2 = (2.25 - 0.56)×10^-6 = 1.69×10^-6

=> x = ?1.69×10^-6 = 1.3×10^-3

Thus speed becomes half of maximum value at a distance of 1.3×10^-3 m = 1.3 mm from the equillibrium position.

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