QUESTION 4 Consider a harmonic oscillator with mass 0.26 kg and spring constant

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QUESTION 4 Consider a harmonic oscillator with mass 0.26 kg and spring constant 175 N/m. If the amplitude is 8.5 cm, what is the speed of the mass at a point which is displaced by 72% of the amplitude off the equilibrium point? (work this out using conservation of energy. Practise how to draw an energy bar chart for this.) Answer with units of m/s. Answer with exactly 3 significant figures, (for example 1.23m/s or 0.345m/s or 0.0876m/s or 12.4m/s)

QUESTION 5 Consider a harmonic oscillator with period 0.07 s. If the amplitude is 8.14 cm, and at a certain time the mass is found to be moving at 1.67 m/s, what is the magnitude of the displacement from the equilibrium position (answer in units of cm). (work this out using conservation of energy. Practise how to draw an energy bar chart for this.) Answer with units of cm. Answer with exactly 3 significant figures (for example 1.23cm or 0.345cm or 0.0876cm or 12.4cm)

Explanation / Answer

4)

apply law of conservation of energy

total energy at the Amplitude = total energy at the 72% of the amplitude

0.5*k*A^2 =(0.5*K*x^2)+(0.5*m*v^2)

k = 175 N/m
A = 8.5 cm = 0.085 m

x = 0.72*0.085 = 0.0612 m

then 0.5*k*A^2 =(0.5*K*x^2)+(0.5*m*v^2)

0.5*175*0.085*0.085= (0.5*175*0.0612*0.0612)+(0.5*0.26*v^2)

v = 1.53 m/s

5) T = 0.07 S

A = 8.14 cm = 0.0814 m

vmax = A*(2*pi/T) = 0.0814*2*3.142/0.07 = 7.307 m/s

use law of conservation of energy

0.5*m*Vmax^2 = (0.5*m*v^2)+(0.5*k*x^2)

mass m = k/w^2 = k/(2*pi/T)^2 = k/(6.284/0.07)^2 = 1.24*10^-4*k

0.5*1.24*10^-4*K*7.3*7.3 = (0.5*1.24*10^-4*k*1.67*1.67) +(0.5*k*x^2)

k cancels on both sides

0.0033 = 0.0001729118 + (0.5*x^2)

x = 0.07908 m= 7.9 cm

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