Question

A simple harmonic oscillator consists of a block (m = 0.50 kg) attached to a spring (k = 128 N/m). The block is pulled a certain distance from the equilibrium position and released at t = 0 s. The block slides on a horizontal frictionless surface about the equilibrium point x = 0 m with a total mechanical energy of 16 J.a) What are the amplitude and phase constant of the oscillation? b) Find the period of the motion and determine the number of oscillations (cycles) the block completes during a 5.5 s interval? c) Find the maximum speed the block will achieve and the time at which it first occurs. d) Find the speed of the block at the instant it is displaced 0.35 m from its equilibrium position.

Answer 1

Answer

given,

k = spring constant = 128 N/m

mass of block = 0.5 Kg

mechanical energy = 16 J

using formula

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

$16 = \dfrac{1}{2}\times 128 \times A^2$

A² = 0.25

A = 0.5 m

b) T = time period of one oscillation

   $T = 2\pi \sqrt{\dfrac{m}{k}}$

   $T = 2\pi \sqrt{\dfrac{0.5}{128}}$  

            T = 0.393 s

number of oscillations = $\dfrac{5.5}{0.393}$

    n = 14

c)  

maximum K E = mechanical energy = 16 J

d)  using conservation of energy

K E + E_{sp} = mechanical energy

$\dfrac{1}{2}mv^2 +\dfrac{1}{2}kx^2 =16$

at x = 0.35  m

$\dfrac{1}{2}\times 0.5\times v^2 +\dfrac{1}{2}\times 128 \times 0.35^2 =16$

v = 5.71 m/s

Answer 2

Answer:

0.5 m

14.00595

8 m/s, 0.0625 s

5.71314 m/s

Explanation:

k = Spring constant = 128 N/m

A = Amplitude

E = Energy in spring = 16 J

Energy in spring is given by

$E=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{2E}{k}}\\\Rightarrow A=\sqrt{\dfrac{2\times 16}{128}}\\\Rightarrow A=0.5\ m$

The amplitude is 0.5 m

Time period is given by

$T=2\pi\sqrt{\dfrac{m}{k}}\\\Rightarrow T=2\pi\sqrt{\dfrac{0.5}{128}}\\\Rightarrow T=0.39269\ s$

Number of oscillations is given by

$N=\dfrac{5.5}{0.39269}\\\Rightarrow N=14.00595$

The number of oscillations is 14.00595

For maximum speed

$\dfrac{1}{2}mv^2=16\\\Rightarrow v=\sqrt{\dfrac{16\times 2}{0.5}}\\\Rightarrow v=8\ m/s$

The maximum speed is 8 m/s

For a distance of 0.5 m which is the amplitude

$Time=\dfrac{Distance}{Speed}\\\Rightarrow Time=\dfrac{0.5}{8}\\\Rightarrow Time=0.0625\ s$

The time taken would be 0.0625 s

The maximum kinetic energy is equal to the mechanical energy

$\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2=16$

At x = 0.35 m

$v=\sqrt{\dfrac{16-\dfrac{1}{2}kx^2}{\dfrac{1}{2}m}}\\\Rightarrow v=\sqrt{\dfrac{16-\dfrac{1}{2}128\times 0.35^2}{\dfrac{1}{2}0.5}}\\\Rightarrow v=5.71314\ m/s$

The speed of the block is 5.71314 m/s