Question

A mass of 1.50 kg is attached to a spring and the system is undergoing simple harmonic oscillations with a frequency of 1.85 Hz and an amplitude of 6.92 cm. What is the speed of the mass when it is 3.34 cm from its equilibrium position?.

Answer 1

find the angular velocity first\[\omega=2\pi f\]and then the spring constant\[k=m \omega^{2}\]after that use the conservation of energy to find the velocity

Answer 2

according to your question :
when the body in equilibrium it will have maximum velocity.
hence maximum velocity =\[Vmax=A \times \omega\]
apply conservation of energy::
\[1\div 2(m \times Vmax ^{2})+0=1\div 2(m \times V ^{2})+1\div 2(m \times \omega ^{2} \times X ^{2})\]
put X=3.34*10^-2, and find V.....

Answer 3

from here\[1/2(m*w^2* ?\]

Answer 4

@mani_jha

Answer 5

It's:
1/2(m*w^2*x^2)
where x is the position of the mass(which in this case would be 3.34 cm)
Did you understand the rest?

Answer 6

what about the k=spring constant?

Answer 7

Since it's not given in the question, it's really not required. Besides, the energy of a particle in SHM is given by (1/2)m*w^2*x^2. As long as you know w, you don't require k.(Only if you solve using conservation of energy)
If you solve using Newton's laws(writing force equations), you would need to find k using:
\[w=\sqrt{k/m}\]