Kvant Physics Problem 24
A body of mass $M$ is initially resting on a stand and attached to a vertical spring of spring constant $k$.
Verified: yes
Verdicts: PASS + PASS
Solve time: 8m17s
Source on kvant.digital
Problem
The stand on which a body rests, suspended from a spring, is set into downward motion with acceleration $a$. At the initial moment the spring is unstretched. After what time will the body lose contact with the stand? To what maximum length will the spring stretch? The mass of the body is $M$, and the spring constant is $k$.
N. I. Goldfarb
Setup and Assumptions
A body of mass $M$ is initially resting on a stand and attached to a vertical spring of spring constant $k$. The stand begins to move downward with constant acceleration $a$. At the initial instant, $t=0$, the spring is unstretched.
The required quantities are the time $t_1$ at which the body loses contact with the stand, and the maximum extension of the spring after separation.
The gravitational acceleration is denoted by $g$.
The spring is assumed ideal and massless. The body is treated as a point mass. Air resistance is neglected. The acceleration $a$ of the stand is constant. Motion is considered in an inertial frame fixed to the Earth. The positive direction is chosen downward.
While the body remains in contact with the stand, the body and stand have the same acceleration. Contact is lost when the normal reaction between the stand and the body becomes zero.
Physical Principles
The solution is based on Newton's second law,
$$\sum F = M a_{\text{body}}.$$
The force exerted by a spring obeys Hooke's law,
$$F_s = kx,$$
where $x$ is the spring extension.
The body loses contact with the stand when the normal force $N$ vanishes,
$$N=0.$$
After separation, the body moves under the action of gravity and the spring force only. Its motion is then governed by
$$M\ddot x = Mg-kx.$$
The equilibrium extension of the spring for a freely suspended mass is
$$x_{\text{eq}}=\frac{Mg}{k}.$$
The total mechanical energy is conserved after separation because only conservative forces act.
Derivation
While the body remains on the stand, its acceleration is equal to the prescribed acceleration of the stand,
$$a_{\text{body}}=a.$$
Let $x$ be the extension of the spring. Since the body and stand move together and the spring is initially unstretched,
$$x=\frac12 at^2.$$
The forces acting on the body are the weight $Mg$, the spring force $-kx$, and the normal force $N$. Taking downward as positive, Newton's second law gives
$$Mg-kx-N=Ma.$$
Hence
$$N=Mg-kx-Ma.$$
The condition for loss of contact is
$$N=0.$$
Substituting into the previous equation,
$$Mg-kx-Ma=0,$$
or
$$kx=M(g-a).$$
The corresponding spring extension at separation is
$$x_1=\frac{M(g-a)}{k}.$$
Since $x=\frac12 at^2$ during contact,
$$\frac12 at_1^2=\frac{M(g-a)}{k}.$$
Solving for $t_1$,
$$t_1=\sqrt{\frac{2M(g-a)}{ka}}.$$
This result is meaningful only when $a<g$. If $a\ge g$, the body loses contact immediately because the required normal force would be negative at $t=0$.
The velocity at the instant of separation is
$$v_1=at_1 =\sqrt{\frac{2Ma(g-a)}{k}}.$$
After separation the body obeys
$$M\ddot x=Mg-kx.$$
This is oscillatory motion about
$$x_{\text{eq}}=\frac{Mg}{k}.$$
The angular frequency is
$$\omega=\sqrt{\frac{k}{M}}.$$
At the instant of separation,
$$x_1=\frac{M(g-a)}{k},$$
so the displacement from equilibrium is
$$y_1=x_1-x_{\text{eq}} =-\frac{Ma}{k}.$$
The corresponding velocity is
$$v_1=\sqrt{\frac{2Ma(g-a)}{k}}.$$
The amplitude of oscillation about equilibrium is
$$A=\sqrt{y_1^2+\left(\frac{v_1}{\omega}\right)^2}.$$
Substituting the expressions for $y_1$, $v_1$, and $\omega$,
$$A^2 =\left(\frac{Ma}{k}\right)^2 +\frac{M}{k}\frac{2Ma(g-a)}{k}.$$
Thus
$$A^2 =\frac{M^2}{k^2} \left[a^2+2a(g-a)\right] =\frac{M^2}{k^2}a(2g-a),$$
and
$$A=\frac{M}{k}\sqrt{a(2g-a)}.$$
The greatest spring extension occurs at the lower turning point,
$$x_{\max}=x_{\text{eq}}+A.$$
Hence
$$x_{\max} =\frac{Mg}{k} +\frac{M}{k}\sqrt{a(2g-a)}.$$
Therefore,
$$x_{\max} =\frac{M}{k} \left[g+\sqrt{a(2g-a)}\right].$$
Result
For $a<g$, the body remains in contact with the stand until the spring extension reaches
$$x_1=\frac{M(g-a)}{k}.$$
The time at which contact is lost is
$$\boxed{ t_1=\sqrt{\frac{2M(g-a)}{ka}} }.$$
The maximum spring extension after separation is
$$\boxed{ x_{\max} =\frac{M}{k} \left[g+\sqrt{a(2g-a)}\right] }.$$
No numerical values are provided in the statement, so the answer remains in symbolic form.
For $a\ge g$,
$$\boxed{t_1=0},$$
and the body separates from the stand immediately.
Sanity Checks
The expression
$$t_1=\sqrt{\frac{2M(g-a)}{ka}}$$
has dimensions
$$\sqrt{\frac{\text{kg}\cdot \text{m s}^{-2}} {(\text{N m}^{-1})(\text{m s}^{-2})}} = \sqrt{\text{s}^2} = \text{s},$$
which is correct for a time.
The expression
$$x_{\max} =\frac{M}{k} \left[g+\sqrt{a(2g-a)}\right]$$
has dimensions
$$\frac{\text{kg}}{\text{N m}^{-1}} \cdot \frac{\text{m}}{\text{s}^2} = \text{m},$$
which is correct for a length.
In the limit $a\to g^{-}$,
$$t_1\to 0,$$
so separation occurs immediately. This agrees with the fact that a stand falling with acceleration $g$ cannot press on the body.
In the limit $a\to 0$, the formula for $t_1$ diverges. A stationary stand never moves away from the body, so contact is never lost, which matches the divergence.
For $a=g$,
$$x_{\max} =\frac{Mg}{k} +\frac{Mg}{k} = \frac{2Mg}{k}.$$
The body is released from the natural length of the spring with zero velocity, so it falls through a distance $2Mg/k$ before reaching the lower turning point. This is the standard result for a mass suddenly attached to a vertical spring.
The most common sign error occurs in writing Newton's second law during the contact phase. With downward chosen as positive, the spring force enters as $-kx$, leading to
$$Mg-kx-N=Ma.$$
Changing that sign reverses the condition for separation and corrupts both final results.