Answer to Question #190846 in Mechanics | Relativity for DANI

Question

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A 200g block connected to a light spring for which the force constant is 5N/m is free to oscillate on a horizontal friction less surface. The block is displaced from equilibrium position and released from rest.

1)Determine amplitude, frequency and period.

2)Determine the maximum velocity and maximum acceleration.

3)Find the position, velocity and acceleration when time is one second.

Answer 1

Explanations & Calculations

• The spring is weightless and when a system of this kind is on a horizontal surface, the equilibrium position is just the point at which the spring is about to stretch. (If the length of the spring is $l$, then the distance to the equilibrium point from the fixed point is $l$).
• For this type of situation (horizontal), the amplitude of the following motion depends on how much the initial displacement will be given. Here some data is missing & if we take the initial displacement to be $d$, then the amplitude of the motion will be just $d$.

$\qquad\qquad \begin{aligned} \small A&=\small d \end{aligned}$

• To determine the frequency, apply Newton's second law on the mass for a situation where the system is displaced from its eq. position and now recoiling in.

$\qquad\qquad \begin{aligned} \small \rightarrow F&=\small ma \\ \small -kx&=\small m\ddot{x}\\ \small \ddot{x}&=\small -\frac{k}{m}x \end{aligned}$

• Comparison with the characteristic equation of s.h.m yeilds,

$\qquad\qquad \begin{aligned} \small \omega^2&=\small \frac{k}{m}\\ \small \omega&=\small \sqrt{\frac{k}{m}}\\ &=\small \bold{5\,rad s^{-1}} \end{aligned}$ the frequancy

$\\ \qquad\qquad \begin{aligned} \small \omega&=\small \frac{2\pi}{T}=5\\ \small T&=\small \bold{1.26\,s} \end{aligned}$ the period

2.

• The maximum velocity is experienced at the equilibrium position and it is given by the equation $\small V_{max}=A\omega$. Then,

$\qquad\qquad \begin{aligned} \small V_{max}&=\small d\times 5\\ &=\small \bold{5d} \,ms^{-1} \end{aligned}$

• Maximum acceleration is experienced at the amplitude, according to the characteristic equation.

$\qquad\qquad \begin{aligned} \small |a_{max}|&=\small \omega^2x_{max}=\omega^2A=5^2\times d= \bold{25d} \end{aligned}$

3.

• Position after 1s can be calculated according to the general equation as follows,

$\qquad\qquad \begin{aligned} \small x&=\small A\sin\omega t=d\sin(5\times1)\\ &=\small \bold{0.087d}\,\,m \end{aligned}$

• The velocity can be found by derivating the position equation once w.r.t time

$\qquad\qquad \begin{aligned} \small V&=\small \dot{x}=A\omega\cos(\omega t)=5d\cos(5\times 1)\\ &=\small \bold{4.98d}\,\,ms^{-1} \end{aligned}$

• Since the position after 1s is known by now, the acceleration after 1s can be found by the characteristic equation

$\qquad\qquad \begin{aligned} \small a&=\small \omega^2x=5^2\times0.087d\\ &=\small \bold{2.175d}\,\,ms^{-2} \end{aligned}$

• For all those, the initial displacement needs to be known.