Finding Mass in SHM with Known Spring Constant and Velocity

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SUMMARY

The discussion focuses on calculating the mass of a block undergoing simple harmonic motion (SHM) attached to a spring with a spring constant of 9.00 N/m. The block's velocity is measured at +30.0 cm/s when it is halfway between its equilibrium position and maximum displacement. The correct mass is determined to be 0.750 kg using the relationship between velocity, angular frequency, and mass derived from circular motion principles.

PREREQUISITES
  • Understanding of simple harmonic motion (SHM)
  • Knowledge of spring constants and Hooke's Law
  • Familiarity with circular motion concepts
  • Ability to apply kinematic equations in physics
NEXT STEPS
  • Study the derivation of angular frequency in SHM
  • Learn about the relationship between mass, spring constant, and frequency
  • Explore the application of energy conservation in SHM
  • Investigate the effects of damping on simple harmonic motion
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Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators looking for examples of SHM calculations.

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Homework Statement



A block of unknown mass is attached to a spring with a spring constant 9.00 N/m and
undergoes simple harmonic motion, along the x axis, with an amplitude of 10.0 cm. When the
mass is halfway between its equilibrium position and its maximum positive displacement, its
velocity is measured to be +30.0 cm/s. Calculate
(a) the mass of the block,

Homework Equations



??

The Attempt at a Solution



I tried using the x= A cos (ωt +∅) but i can't figure it out without time. HELP?? I know the answer is suposed to be m= 0.750kg but i don't understand how to get there.
 
Physics news on Phys.org
What is cos(ωt +∅) when the mass is halfway between its equilibrium position and its maximum positive displacement?

What is the formula for velocity as function of t?
How is ω related to mass?

ehild
 
Since OP not interested, i find the problem interesting since i just started on SHM.

Using circular motion method.
Since position is halfway between its equilibrium position and its maximum positive displacement,
θ=60°
VmaxCos(90-θ)=0.3 m/s
Vmax=0.3x 2/√3

F=mv2/r
m=0.9 x 0.1 x 3/(0.36)
m=0.75 kg
 

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