A partridge of mass 4.98 is suspended from a pear tree by an ideal spring of neg

Question

A partridge of mass 4.98 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 4.18 .

Part A
What is its speed as it passes through the equilibrium position?
m/s





Part B
What is its acceleration when it is 0.050 m above the equilibrium position?






Part 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?
s





Part D
The motion of the partridge is stopped, and then it is removed from the spring. How much does the spring shorten?
m

Explanation / Answer

Let positive direction means downward, and let positive displacement is below equilibrium point and negative above it. 1) Angular frequency: ? = 2pf = 2p / T ? = 2p / 4.18 = 1.502 s^-1 v= -A ? sin(? t) where A=0.1m is amplitude For the first pass through equilibrium position, t=T/4 and ?t=2pt /T= p/2, thus v= vmax = -A ? v = -0.1 * 1.502 = -0.1502m/s Note: to find formula for vmax one possible approach is to find maximal potential energy Epmax=kA^2/2, where k is spring constant, k= m?^2 = m (2 p/T)^2 and to apply the law of conservation of energy. In equilibrium position potential energy is converted to kinetic: m vmax^2/2 = kA^2/2 vmax=A sqrt(k/m) = A? 2) a = -?^2 y a = -1.502^2 * (-5x10^-2) = 11.280x10^-2 m/s^2 positive sign corresponds to downward direction

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