A particle with mass 2.69 kg oscillates horizontally at the end of a horizontal

Question

A particle with mass 2.69 kg oscillates horizontally at the end of a horizontal spring. A student measures an amplitude of 0.933 m and a duration of 131 s for 66 cycles of oscillation. Find the frequency, f, the speed at the equilibrium position, Vmax, the spring constant, k, the potential energy at an endpoint, Umax, the potential energy when the particle is located 30.7% of the amplitude away from the equilibrium position, U, and the kinetic energy, K, and the speed, v, at the same position. Number f= Hz Number m/ s Number N/ m Number (Scroll down for more answer blanks.) Umax= 10

Explanation / Answer

Time period = time taken for one cycle = 131/ 66 = 1.984848485 seconds

Frequency = 1 / time period = f = 66 / 131 = 0.503816793 Hz

frequncy f = (1/2) ( k/m )1/2

0.503816793 x 2 = ( k/2.69 )1/2

k/2.69 = 10.0208605

k = 26.95611475 N/m

So, maximum potential energy stored in the spring = 0.5 k A2   where A is the Amplitude

E = 0.5 x 26.95611475 x 0.9332 =11.73250069 Joules

so, Umax = E = 11.73250069 J

When the mass is at the equilibrium position, this energy will be converted into the kinetic eneryg of the mass with velocity vmax

E = 0.5mvmax2

11.73250069 = 0.5 x 2.69 x vmax2

vmax = 2.9534808 m/s

When particle is at 30.7 % of it's amplitude, deflection of spring = 0.307 x 0.933= 0.286431 m

Potential energy at this point = 0.5 k 2 = 0.5 x 26.95611475 x  0.2864312 =1.105776457 J

U = 1.105776457 J

So, out of the total energy E, if this the potnetial energy at that point, the remaining will be the kinetic energy

KE = E - 1.105776457 = 11.73250069 - 1.105776457 = 10.62672423 J

K = 10.62672423 J

if v be the speed to give this kinetic energy , then KE = 0.5mv2

so v2 = 2KE /m = 2 x 10.62672423 / 2.69 = 7.90091021

v = 2.810855779 m/s

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