Apartridge 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?

Part a)

Part b)

Part c)

Part d)

Explanation:

Time period of the spring is given as

now we know that

m = 4.90 kg

so we have now

Part a)

Amplitude = 0.100

So speed at mean position is given as

Part b)

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

Part c)

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

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

gets less

the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever