Chapter 11, Problem 31SP

A 300-g mass at the end of a Hookean spring vibrates up and down in such a way that it is 2.0 cm above the tabletop at its lowest point and 16 cm above at its highest point. Its period is 4.0 s. Determine (a) the amplitude of vibration, (b) the spring constant, (c) the speed and acceleration of the mass when it is 9 cm above the tabletop, (d) the speed and acceleration of the mass when it is 12 cm above the tabletop.

To determine

The amplitude of the spring vibration when the mass of $\text{300 g}$ is attached to a Hookean spring and is $\text{2 cm}$ above the tabletop at its lowest point and $\text{16 cm}$ above its highest point.

Answer to Problem 31SP

Solution:

$7\text{ cm}$

Explanation of Solution

Given data:

The mass of $\text{300 g}$ is attached to a Hookean spring and is $\text{2 cm}$ above the tabletop at its lowest point and $\text{16 cm}$ above its highest point.

Time period is $4.0\text{ s}$.

Formula used:

The amplitude of an hormonic motion is the half of the distance between the end points of the motion.

Mathematically,

$x_0=\frac{x_h-x_l}{2}$

Here, $x_0$ is the amplitude of the motion, $x_h$ is the highest position of the motion, and $x_l$ is the lowest position of the motion.

Explanation:

The amplitude is the maximum displacement of the spring from the equilibrium position.

Recall the expression for the amplitude:

$x_0=\frac{x_h-x_l}{2}$

Substitute $16\text{ cm}$ for $x_h$ and $2\text{ cm}$ for $x_l$

$\begin{array}{c}{x}_0=\frac{16\text{ cm}-2\text{ cm}}{2}\\ =\frac{14\text{ cm}}{2}\\ =7\text{ cm}\end{array}$

Conclusion:

Therefore, the amplitude of the vibration is $7\text{ cm}$.

To determine

The spring constant of the spring when the mass of $\text{300 g}$ is attached to a Hookean spring and is $\text{2 cm}$ above the tabletop at its lowest point and $\text{16 cm}$ above its highest point

Answer to Problem 31SP

Solution:

$\text{0}\text{.74 N/m}$

Explanation of Solution

Given data:

The mass of $\text{300 g}$ is attached to a Hookean spring and is stretched by $\text{2 cm}$ above the tabletop at its lowest point and $\text{16 cm}$ above its highest point.

Time period is $4.0\text{ s}$.

Formula used:

The formula for time period of a simple spring mass system in SHM is expressed as

$T=2\pi \sqrt{\frac{m}{k}}$

Here, $k$ is the elastic spring constant, $m$ is the value of attached mass, and $T$ is the time period of vibration of the system.

Explanation:

Consider the expression for time period of the spring–mass system for the $\text{300 g}$ mass:

$T=2\pi \sqrt{\frac{m}{k}}$

Rearrange for $k$

$k=m{\left(\frac{2\pi }{T}\right)}^2$

Substitute $\text{300 g}$ for $m$ and $4\text{ s}$ for $T$

$\begin{array}{c}k=300\text{ g}\left(\frac{1\text{ kg}}{1000\text{ g}}\right){\left(\frac{2\left(3.14\right)}{4\text{ s}}\right)}^2\\ =\left(0.3\right)\left(2.46\right)\\ =0.74\text{ N/m}\end{array}$

Conclusion:

Therefore, the spring constant is $\text{0}\text{.74 N/m}$.

To determine

The speed and magnitude of the acceleration of the mass when the mass of $\text{300 g}$ is attached to a Hookean spring and is $\text{9 cm}$ above the tabletop.

Answer to Problem 31SP

Solution:

$0.1098\text{ m/s}$ and zero

Explanation of Solution

Given data:

The mass of $\text{300 g}$ is attached to a Hookean spring and is stretched by $\text{2 cm}$ above the tabletop at its lowest point and $\text{16 cm}$ above its highest point.

Time period is $4.0\text{ s}$.

Formula used:

The expression for velocity of mass in SHM at a location $x$ can be written as

$v=\sqrt{\left(x_0^2-x^2\right)\left(\frac{k}{m}\right)}$

Here, $v$ is the velocity of mass at a location $x$, $x_0$ is the amplitude or the maximum displacement of the mass from the mean position, $k$ is the spring constant, and $m$ is the mass that attached to the spring.

The expression for magnitude of acceleration of a mass in terms of displacement and time period can be written as

$a=\frac{4{\pi }^2}{T^2}x$

Here, $x$ is the displacement of the mass from the mean position, $T$ is the time period, and $a$ is the acceleration of the body.

Explanation:

Consider the expression for velocity of the mass $m$:

$v=\sqrt{\left(x_0^2-x^2\right)\left(\frac{k}{m}\right)}$

Substitute $7\text{ cm}$ for $x_0$, $0$ for $x$, $\text{300 g}$ for $m$, and $\text{0}\text{.74 N/m}$ for $k$

$\begin{array}{c}v=\sqrt{\left(\left(7\text{ cm}\left(\frac{1\text{ m}}{100\text{ cm}}\right)\right)-{\left(0\right)}^2\right)\left(\frac{0.74\text{ N/m}}{300\text{ g}\left(\frac{1\text{ kg}}{100\text{0 g}}\right)}\right)}\\ =\sqrt{4.9\times {10}^{-3}\times 2.466}\\ =\sqrt{0.012}\\ =0.1098\text{ m/s}\end{array}$

Understand that the given position of the body hung to spring will be at equilibrium position. Therefore, the value of $x$ is zero.

Recall the expression for $a$

$a=\frac{4{\pi }^2}{T^2}x$

Substitute $\text{0 cm}$ for $x$ and $4\text{ s}$ for $T$

$\begin{array}{c}a=\frac{4{\pi }^2}{T^2}\left(0\right)\\ =0\end{array}$

Conclusion:

The speed is $0.1098\text{ m/s}$ and acceleration of the mass at the given position is zero.

To determine

The speed and magnitude of the acceleration of the mass when the mass of $\text{300 g}$ is attached to a Hookean spring and is stretched by $\text{12 cm}$ above the tabletop.

Answer to Problem 31SP

Solution:

$0.099\text{ m/s}$ and $0.074{\text{ m/s}}^2$

Explanation of Solution

Given data:

The mass of $\text{300 g}$ is attached to a Hookean spring and is stretched by $\text{2 cm}$ above the tabletop at its lowest point and $\text{16 cm}$ above its highest point.

Time period is $4.0\text{ s}$.

Formula used:

The expression for magnitude of acceleration of a mass in terms of displacement and time period can be written as

$a=\frac{4{\pi }^2}{T^2}x$

Here, $x$ is the displacement of the mass from the mean position, $T$ is the time period, and $a$ is the acceleration of the body.

The expression for velocity of mass in SHM at a location $x$ can be written as

$v=\sqrt{\left(x_0^2-x^2\right)\left(\frac{k}{m}\right)}$

Here, $v$ is the velocity of mass at a location $x$, $x_0$ is the amplitude or the maximum displacement of the mass from the mean position, $k$ is the spring constant, and $m$ is the mass that attached to the spring.

Explanation:

Consider the expression for velocity of the mass $m$:

$v=\sqrt{\left(x_0^2-x^2\right)\left(\frac{k}{m}\right)}$

Substitute $7\text{ cm}$ for $x_2$, $3\text{ cm}$ for $x$, $\text{300 g}$ for $m$, and $\text{0}\text{.74 N/m}$ for $k$

$\begin{array}{c}v=\sqrt{\left(\left(7\text{ cm}\left(\frac{1\text{ m}}{100\text{ cm}}\right)\right)-{\left(3\text{ cm}\left(\frac{1\text{ m}}{100\text{ cm}}\right)\right)}^2\right)\left(\frac{0.74\text{ N/m}}{300\text{ g}\left(\frac{1\text{ kg}}{100\text{0 g}}\right)}\right)}\\ =\sqrt{4\times {10}^{-3}\times 2.466}\\ =\sqrt{9.864\times {10}^{-3}}\\ =0.099\text{ m/s}\end{array}$

Recall the expression for $a$

$a=\frac{4{\pi }^2}{T^2}x$

Understand that the position of the body hung to spring will be at $3\text{ cm}$ above the equilibrium position because the equilibrium position is $9\text{ cm}$ above the tabletop. Therefore, the value of $x$ is $3\text{ cm}$.

Substitute $3\text{ cm}$ for $x$ and $4\text{ s}$ for $T$

$\begin{array}{c}a=\frac{4{\left(3.14\right)}^2}{{\left(4\text{ s}\right)}^2}\left(3\text{ cm}\left(\frac{1\text{ m}}{100\text{ cm}}\right)\right)\\ =0.074{\text{ m/s}}^2\end{array}$

Conclusion:

The velocity $12\text{ cm}$ above the tabletop is $0.099\text{ m/s}$ and the acceleration is $\text{0}{\text{.074 m/s}}^2$.