Q7.7 Consolidation exercise (K)
One end of a spring is fastened to a rigid support. A calibrated spring balance attached to the free end is used to determine that the restoring force is proportional to displacement with a force of 4 N giving a displacement of 0.02 m. A 2 kg object is then attached to the end of the spring, allowed to come to equilibrium, then pulled a distance of 0.04 m, and released.- Find the spring constant k.
- Find the angular frequency, the frequency and the period of vibration.
- Find the maximum velocity and acceleration.
- How much time is required for the object to move halfway back to the centre from its initial position?
Solution
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Solution
- The spring constant k prescribes the restoring force exerted by the spring, per unit displacement. The data provided shows that k×0.02=4 implying k=200 Nm-1
- Writng Newton’s second law for the mass, moving under
the action of the spring we have
ma=-kximplyingThe value of the angular frequency is thus
![\[ \ddot{x} = -\omega^2 x \hspace{0.5cm}\mbox{\rm with}\hspace{0.5cm} \omega = \sqrt{\frac{k}{m}} \]](mastermathpng-0.png)
The period follows as
while the frequency is![\[ T = \frac{2\pi}{\omega}= 0.628s \]](mastermathpng-2.png)
![\[ f=\frac{1}{T} = 1.59\,Hz \]](mastermathpng-3.png)
- The general equation for the displacement (remember it!)
x=xmcos(ωt+φ)implies the equation for the velocity (which, therefore, you don’t need to remember...as long as you can differentiate confidently!):which has its maximum when the sine function is unity. The maximum speed is thus
vmax=ωxm=0.4ms-1Similarly, the accelerationhas maximum value
amax=ω2xm=4ms-2 - The initial conditions are x(t=0)=xm and v(t=0)=0, since the
particle is released from rest. The specific form of the general solution
we need to match these initial conditions is
x=xmcos(ωt)The time required to reach the point x=xm/2 is the solution ofimplying
so t=π/30 s=T/6. Contrast this with the time it takes to reach the equilibrium point x=0, which is T/4. So it takes T/6 to travel from x=xm to x=xm/2 but only a further T/12 to travel from x=xm/2 to x=0. The distances covered in the two intervals is the same (namely xm/2); but the second one takes less time because the average speed is bigger.![\[ \omega t = \cos^{-1} \left(\frac{1}{2} \right) = \frac{\pi}{3} \]](mastermathpng-7.png)