Question

A 2.00 kg , horizontal, uniform tray is attached to a vertical ideal spring of force constant 190 N/m and a 295 g metal ball is in the tray. The spring is below the tray, so it can oscillate up-and-down. The tray is then pushed down 17.4 cm below its equilibrium point (call this point A) and released from rest.
(a) How high above point A will the tray be when the metal ball leaves the tray?(b) How much time elapses between releasing the system at point A and the ball leaving the tray?(c) How fast is the ball moving just as it leaves the tray?

Answer 1

Answer:

a) $d=29.24\ cm$

b) $t=0.09\ s$

c) $v_y=1.16\ m.s^{-1}$

Explanation:

Given:

• mass of the tray, $M=2\ kg$

• mass of the ball, $m=0.295\ kg$

• spring constant, $k=190\ N.m^{-1}$

• amplitude of spring oscillation, $A= 0.174\ m$

(a)

Now the ball will leave the tray at the highest oscillation position from the equilibrium where the spring force becomes equal to the combined weights.

mathematically:

$k.x=(m+M)g$

where:

x = maximum position of ball and tray from the initial mean position.

$190\ x =9.8(2.295)$

$x=0.1184\ m$

$x=11.84\ cm$

∴Distance from point A which is fully compressed position:

$d=x+A$

$d=11.84+17.4$

$d=29.24\ cm$

(b)

Simple harmonic motion as a function of time is given as:

$y=A.cos\ (\omega.t-\phi)$ ....................................(1)

where:

$\omega=$ angular frequency

t = time

$\phi=$ phase angle= 0 (initially at t=0 the spring is fully compressed)

y = displacement

Now from equation (1):

$t=\frac{1}{\omega} \cos^{-1}(\frac{y}{A} )$ ...........................(2)

we know

$\omega = \sqrt{\frac{k}{m'} }$

$\omega = \sqrt{\frac{k}{M+m} }$

$\omega = \sqrt{\frac{190}{2.295} }$

$\omega=9.1 Hz$

Now, putting reaspective values in eq. (2)

$t=\frac{1}{9.1}\times \cos^{-1}(\frac{11.84}{17.4} )$

$t=0.09\ s$

(c)

We have the equation for total mechanical energy of SHM as:

$E=\frac{1}{2} m'.v_y^2+\frac{1}{2} k.y^2=\frac{1}{2} k.A^2$

$v_y^2= k\frac{A^2- y^2}{M+m}$

$v_y^2= 190\times\frac{0.174^2- 0.1184^2}{2.295}$

$v_y=1.16\ m.s^{-1}$

Answer 2

For the horizontal, uniform tray which is attached to a vertical ideal spring and a metal ball is in the tray, the values of height, time and speed is,

• The height above point A will the tray be when the metal ball leaves the tray is 29.24 cm.
• (b) The total time elapses between releasing the system at point A and the ball leaving the tray is 0.09 seconds.

• (c) The speed of the ball moving just as it leaves the tray is 1.16 m/s.

What is spring constant?

Spring constant is nothing but the stiffness of the spring. Spring constant is the ratio of force applied on the spring to the displacement of the spring.

A 2.00 kg , horizontal, uniform tray is attached to a vertical ideal spring of force constant 190 N/m and a 295 g metal ball is in the tray.

• (a) The height above point A will the tray be when the metal ball leaves the tray-

The total mass of the tray and ball is 2.295 kg (2kg+0.295 kg). The force on the spring is equal to the product of spring constant and displacement. Thus,

$F=kx$

By second law,

$ma=kx\\(2.295)9.8=190x\\x=11.84\rm \;cm$

As, the tray is then pushed down 17.4 cm below its equilibrium point (call this point A) and released from rest. Thus, the position of the tray above point A is,

$h=11.84+17.4\\h=29.24\rm \; cm$

• (b) The total time elapses between releasing the system at point A and the ball leaving the tray

The formula for simple harmonic function for spring can be given as,

$y=A\cos(\sqrt{\dfrac{k}{m}}t-\phi)$

Here, (k) is spring constant, (A) is amplitude, (t) is time taken and (y) is displacement. Plug in the values keeping initial phase angle zero,

$11.84=17.4\cos(\sqrt\dfrac{190}{2.295}}t-0)\\t=0.09\rm s$

• (c) The speed of the ball moving just as it leaves the tray

For simple harmonic motion of spring, the total mechanical energy can be given as,

$\dfrac{1}{2}mv_y^2+\dfrac{1}{2}ky^2=\dfrac{1}{2}kA^2\\\dfrac{1}{2}(2.295)v_y^2+\dfrac{1}{2}(190)(0.1184^2)^2=\dfrac{1}{2}(190)(0.174)^2\\v_y^2=1.16\rm m/s$

Thus, for the horizontal, uniform tray which is attached to a vertical ideal spring and a metal ball is in the tray, the values of height, time and speed is,

• The height above point A will the tray be when the metal ball leaves the tray is 29.24 cm.
• (b) The total time elapses between releasing the system at point A and the ball leaving the tray is 0.09 seconds.

• (c) The speed of the ball moving just as it leaves the tray is 1.16 m/s.

Learn more about the spring constant here;

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