Question

A 0.500-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 10.0 cm. Calculate the maximum value of its (a) speed and (b) acceleration, (c) the speed and (d) the acceleration when the object is 6.00 cm from the equilibrium position, and (e) the time interval required for the object to move from x = 0 to x = 8.00 cm.
a. 40cm/s,
b. 160 cm/s2,
c. 32cm/s,
d. -96cm/s2,
e. 0.232s

Answer 1

Answer:

a) $v_{max}=0.4\ m.s^{-1}$

b) $a_{max}=1.6\ m.s^{-2}$

c) $v_x=0.32\ m.s^{-1}$

d) $a_x=0.96\ m.s^{-1}$

e) $\Delta t=0.232\ s$

Explanation:

Given:

mass of the object attached to the spring, $m=0.5\ kg$

spring constant of the given spring, $k=8\ N.m^{-1}$

amplitude of vibration, $A=0.1\ m$

a)

Now, maximum velocity is obtained at the maximum Kinetic energy and the maximum kinetic energy is obtained when the whole spring potential energy is transformed.

Max. spring potential energy:

$PE_s=\frac{1}{2} .k.A^2$

$PE_s=0.5\times 8\times 0.1^2$

$PE_s=0.04\ J$

When this whole spring potential is converted into kinetic energy:

$KE_{max}=0.04\ J$

$\frac{1}{2}.m.v_{max}^2=0.04$

$0.5\times 0.5\times v_{max}^2=0.04$

$v_{max}=0.4\ m.s^{-1}$

b)

Max. Force of spring on the mass:

$F_{max}=k.A$

$F_{max}=8\times 0.1$

$F_{max}=0.8\ N$

Now acceleration:

$a_{max}=\frac{F_{max}}{m}$

$a_{max}=\frac{0.8}{0.5}$

$a_{max}=1.6\ m.s^{-2}$

c)

Kinetic energy when the displacement is, $\Delta x=0.06\ m$:

$KE_x=PE_s-PE_x$

$\frac{1}{2} .m.v_x^2=PE_s-\frac{1}{2} .k.\Delta x^2$

$\frac{1}{2}\times 0.5\times v_x^2=0.04-\frac{1}{2} \times 8\times 0.06^2$

$v_x=0.32\ m.s^{-1}$

d)

Spring force on the mass at the given position, $\Delta x=0.06\ m$:

$F=k.\Delta x$

$F=8\times 0.06$

$F=0.48\ N$

therefore acceleration:

$a_x=\frac{F}{m}$

$a_x=\frac{0.48}{0.5}$

$a_x=0.96\ m.s^{-1}$

e)

Frequency of oscillation:

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

$\omega=\sqrt{\frac{8}{0.5} }$

$\omega=4\ rad.s^{-1}$

So the wave equation is:

$x=A.\sin\ (\omega.t)$

where x = position of the oscillating mass

put x=0

$0=0.1\times \sin\ (4t)$

$t=0\ s$

Now put x=0.08

$0.08=0.1\times \sin\ (4t)$

$t=0.232\ s$

So, the time taken in going from point x = 0 cm to x = 8 cm is:

$\Delta t=0.232\ s$