A 250 g object hangs from a spring and oscillates with an amplitude of 5.42 cm. The spring constant is 48.0 N/m. b. What is the maximum speed of the object? c. At what position will the maximum speed occur?

Question

A 250 g object hangs from a spring and oscillates with an amplitude of 5.42 cm.  The spring constant is 48.0 N/m.  b. What is the maximum speed of the object?  c. At what position will the maximum speed occur?

festinger · Answer

Amplitude is the maximum displacement from the equilbrium position.
System is a spring, assume it obey's hookes law:
$$F=-kx$$and
$$U_{elastic}=\frac{1}{2}kx^{2}$$

At any point the energy is:
$$\frac{1}{2}kx^{2}+\frac{1}{2}mv^{2}= Constant$$

By conservation of energy,
$$U_{max}=KE_{max}$$
This occurs when U is 0, and this occurs when x=0 (can you see why?)
Thus to find maximum speed simple equate (after cancelling the 1/2) $$kA^{2}=mv^{2}$$ where A is the amplitude.

anonymous · Answer

You can use $$E \max = 1/2 mw ^{2}A ^{2} = I/2 m V ^{2}$$,
SO You have $$V = w A$$
also from F=$$-kx =m a$$ i.e$$w ^{2}=k/m$$
$$w=\sqrt{k/m}$$=$$\sqrt{192}=13.9 s ^{-1}$$
so, V max =13.9 * 5.42=75.338  m/s----occurs at the mean position i.e  at unstratched position