Question

A 160 g block attached to a spring with spring constant 2.6 N/m oscillates horizontally on a frictionless table. Its velocity is 21 cm/s when x0 = -4.9 cm . A)What is the amplitude of oscillation? B)What is the block's maximum acceleration? C)What is the block's position when the acceleration is maximum? D)What is the speed of the block when x1 = 3.3 cm ?

Answer 1

A) 0.071 m (7.1 cm)

The table is frictionless, so the mechanical energy of the system is conserved, and it is the sum of the kinetic energy and the potential energy of the spring:

$E=K+U=\frac{1}{2}mv^2+\frac{1}{2}kx^2$

Where

m = 160 g = 0.16 kg is the mass of the block

v = 21 cm/s = 0.21 m/s is the velocity of the block

k = 2.6 N/m is the spring constant

x = -4.9 cm = -0.049 m is the displacement of the block with respect to the equilibrium position

Substituting into the formula, we find:

$E=\frac{1}{2}(0.16 kg)(0.21 m/s)^2+\frac{1}{2}(2.6 N/m)(-0.049 m)^2=0.0066 J$

This is the mechanical energy, and it is constant at any moment of the motion. We also known that the amplitude of the oscillation corresponds to the maximum displacement $A$, and when the system is at maximum displacement, the velocity is zero, and all the energy is elastic potential energy:

$E=\frac{1}{2}kA^2$

From which we find the amplitude:

$A=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(0.0066 J)}{2.6 N/m}}=0.071 m$

B) 1.15 m/s^2

The angular frequency of the system is given by

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{2.6 N/m}{0.16 kg}}=4.03 rad/s$

And the maximum acceleration is given by:

$a_{max}=\omega^2 A=(4.03 rad/s)^2 (0.071 m)=1.15 m/s^2$

C) x = 0.071 m (7.1 cm)

In a simple harmonic motion, the acceleration of the block is related to the displacement x(t) by the formula

$a(t)=-\omega^2 x(t)$

Therefore, we see that the acceleration is proportional to the displacement: this means that when the acceleration is maximum, the position x(t) is also maximum, so the block is located at maximum displacement, which corresponds to the amplitude:

$x=0.071 m=7.1 cm$

D) 0.25 m/s (25 cm/s)

Again, we can solve the problem by using the conservation of energy:

$E=K+U=\frac{1}{2}mv^2+\frac{1}{2}kx^2$

where this time we know:

E = 0.0066 J

m = 0.16 kg

v = ?

k = 2.6 N/m

x = 3.3 cm = 0.033 m

Substituting, we find

$v=\sqrt{\frac{2(E-\frac{1}{2}kx^2)}{m}}=0.25 m/s$

Answer 2

a) The amplitude of oscillation is 7.15 centimeters

b) The maximum acceleration of the block is 116.188 centimeters per square second.

c) The block is at x = -7.15 centimeters when the acceleration is a maximum.

d) The speed of the block when x = 3.3 cm is approximately is 25.570 centimeters per second.

How to analyze a system in a simple harmonic motion

In this case we observe a system under a kind of periodic motion known as simple harmonic motion, in which effects from air viscosity and friction are neglected.

The question requires the understanding and application of the principles of energy conservation and properties of simple harmonic motion to determine each case.

a) By principle of energy conservation we know that initial translational kinetic and spring potential energy (K, U), both in joules, is equal to the maximum potential energy (Uₐ), in joules, from which we calculate the amplitude of oscillation (A):

$\frac{1}{2}\cdot m\cdot v^{2}+\frac{1}{2}\cdot k\cdot x^{2} = \frac{1}{2}\cdot k\cdot A^{2}$   (1)

Where:

• k - Spring constant, in newtons per meter.
• m - Mass, in kilograms
• x - Displacement, in meters

Now we clear the amplitude of oscillation in (1) and determine its value:

$A = \sqrt{\frac{m\cdot v^{2}}{k}+x^{2} }$

$A = \sqrt{\frac{(0.160\,kg)\cdot (0.21\,\frac{m}{s} )^{2}}{2.6\,\frac{N}{m} }+(-0.049\,m)^{2} }$

A = 0.0715 m (7.15 cm)

The amplitude of oscillation is 7.15 centimeters. $\blacksquare$

b) An object under simple harmonic motion have the following relationship between position (x(t)), in centimeters, and acceleration (a(t)), in centimeters per square second:

$a(t) = -\frac{k}{m}\cdot x(t)$   (2)

If we know that x(t) = 7.15 cm, k = 2.6 N/m and m = 0.160 kg, then the maximum acceleration is:

$a(t) = -\left(\frac{2.6}{0.160} \right)\cdot (7.15)$

$a(t) = -116.188\,\frac{cm}{s^{2}}$

The maximum acceleration of the block is 116.188 centimeters per square second. $\blacksquare$

c) Given that the acceleration is a vectorial quantity and opposed to the motion, the acceleration is a maximum when x = -A. Hence, the block is at x = -7.15 centimeters when the acceleration is a maximum. $\blacksquare$

d) We use (1) to calculate the speed of the block:

$\frac{1}{2}\cdot m\cdot v^{2}+\frac{1}{2}\cdot k\cdot x^{2} = \frac{1}{2}\cdot k\cdot A^{2}$

$v = \sqrt{\frac{k}{m}\cdot (A^{2}-x^{2}) }$  

$v = \sqrt{\left(\frac{2.6\,\frac{N}{m} }{0.16\,kg} \right)\cdot [(0.0715\,m)^{2}-(0.033\,m)^{2}]}$

v ≈ 0.256 m/s

The speed of the block when x = 3.3 cm is approximately is 25.570 centimeters per second. $\blacksquare$

To learn more on simple harmonic motion, we kindly invite to check this verified question: https://brainly.com/question/17315536