Chapter 11, Problem 32SP

A coiled Hookean spring is stretched 10 cm when a 1.5-kg body is hung from it. Suppose instead that a 4.0-kg mass hangs from the spring and is set into vibration with an amplitude of 12 cm. Find (a) the force constant of the spring, (b) the maximum restoring force acting on the vibrating body, (c) the period of vibration, (d) the maximum speed and the maximum acceleration of the vibrating object, and (e) the speed and acceleration when the displacement is 9 cm.

To determine

The force constant of the spring when the mass of $1.5\text{kg}$ causing a stretchof $10\text{ cm}$ in a spring is replaced by the mass of $4\text{kg}$ and is set into a vibration with an amplitude of $12\text{ cm}$.

Answer to Problem 32SP

Solution: $\text{0}\text{.15 kN/m}$

Explanation of Solution

Given data:

The mass of $1.5\text{kg}$ is attached to the springcausing a stretchof $10\text{ cm}$.

The $1.5\text{kg}$ mass is replaced by the $4\text{kg}$ mass and is set into a vibration with an amplitude of $12\text{ cm}$.

Formula used:

The restoring force in a spring is expressed as,

$F=-kx$

Here, $k$ is the elastic spring constant, $F$ is the restoring force, and. $x$ is the deflection or change in length of the spring on application of force $F$.

The weight of an object is expressed as,

$W=mg$

Here, $W$ is the weight of the object and $m$ is the mass of the object.

Explanation:

Consider the expression for restoring force in the spring for $1.5\text{ kg}$ mass

$F_1=-k{x}_1$

Here, $x_1$ is the elongation of spring due to $1.5\text{ kg}$ mass and $F_1$ is the restoring force in the spring due to $1.5\text{ kg}$ mass.

Understand that the negative sign shows that the direction of restoring force is opposite to that of the elongation of spring. Consider the magnitude of restoring force

$F_1=k{x}_1$

Consider the expression for weight of the $1.5\text{ kg}$ mass

$W_1=m_{1}g$

Here, $m_1$ is the mass, $W_1$ is the weight of the $1.5\text{ kg}$ mass, and g is the acceleration due to gravity.

$W_1=m_{1}g$

Understand that the weight of mass is equal to the restoring force on the $1.5\text{ kg}$ mass since the system was in equilibrium.

$W_1=F_1$

Substitute $k{x}_1$ for $F_1$ and $m_{1}g$ for $W_1$

$\begin{array}{l}{m}_{1}g=k{x}_1\\ k=\frac{m_{1}g}{x_1}\end{array}$

Understand that the standard value of $g$ is $9.81{\text{ m/s}}^2$. Therefore, substitute $9.81{\text{ m/s}}^2$ for $g$, $1.5\text{ kg}$ for $m_1$, and $10\text{ cm}$ for $x_1$

$\begin{array}{c}k=\frac{\left(1.5\text{ kg}\right)\left(9.81{\text{ m/s}}^2\right)}{10\text{ cm}}\\ =\frac{\left(1.5\text{ kg}\right)\left(9.81{\text{ m/s}}^2\right)}{10\text{ cm}\left(\frac{0.01\text{ m}}{1\text{ cm}}\right)}\\ =\frac{\left(1.5\right)\left(9.81\right)}{0.1}\\ =147.1\text{ N/m}\end{array}$

Further solve as,

$\begin{array}{c}k=\left(147.1\text{ N/m}\right)\left(\frac{0.001\text{kN}}{1\text{ N}}\right)\\ =0.1471\text{kN/m}\\ \approx \text{0}\text{.15 kN/m}\end{array}$

Conclusion:

The force constant for the spring is $\text{0}\text{.15 kN/m}$.

To determine

The maximum restoring force acting on the vibrating body when the mass of $1.5\text{kg}$ causing a stretch of $10\text{ cm}$ in a spring is replaced by the mass of $4\text{kg}$ and is set into a vibration with an amplitude of $12\text{ cm}$.

Answer to Problem 32SP

Solution: $18\text{ N}$

Explanation of Solution

Given data:

The mass of $1.5\text{ kg}$ is attached to the springcausing a stretch of $10\text{ cm}$.

The $1.5\text{kg}$ mass is replaced by the $4\text{kg}$ mass and is set into a vibration with an amplitude of $12\text{ cm}$.

Formula used:

The restoring force in a spring is expressed as,

$F=-kx$

Here, $k$ is the elastic spring constant, $F$ is the restoring force, and $x$ is the deflection or change in length of the spring on application of force $F$.

Explanation:

Consider the expression for restoring force on the mass

$F_2=-k{x}_2$

Here, $F_2$ is the maximum restoring force and $x_2$ is the amplitude of vibration of the $4\text{ kg}$ mass.

Substitute $\text{0}\text{.15 kN/m}$ for $k$ and $12\text{cm}$ for $x_2$

$\begin{array}{c}{F}_2=-\left(\text{0}\text{.15 kN/m}\right)\left(12\text{cm}\right)\\ =-\left(\text{0}\text{.15 kN/m}\right)\left(\frac{1000\text{ N}}{1\text{kN}}\right)\left(12\text{cm}\right)\left(\frac{0.01\text{m}}{1\text{ cm}}\right)\\ =-\left(150\text{N/m}\right)\left(0.12\right)\\ =-18\text{ N}\end{array}$

The negative sign indicates that the restoring force acts in the direction opposite to the elongation of the spring.

Conclusion:

Therefore, themaximum restoring forceacting on the vibrating body is $18\text{ N}$.

To determine

The period of vibration of the vibrating body when the mass of $1.5\text{kg}$ causing a stretch of $10\text{ cm}$ in a spring is replaced by the mass of $4\text{kg}$ and is set into a vibration with an amplitude of $12\text{ cm}$.

Answer to Problem 32SP

Solution: $1.0\text{ s}$

Explanation of Solution

Given data:

The mass of $1.5\text{kg}$ is attached to the springcausinga stretch of $10\text{ cm}$.

The $1.5\text{kg}$ mass is replaced by the $4\text{kg}$ mass and is set into a vibration with an amplitude of $12\text{ cm}$.

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 $4\text{ kg}$ mass

$T_2=2\pi \sqrt{\frac{m_2}{k}}$

Here, $m_2$ is the mass of vibrating body, $T_2$ is the time period of vibration for the mass $m_2$.

Substitute $4\text{ kg}$ for $m_2$ and $\text{0}\text{.15 kN/m}$ for $k$

$\begin{array}{c}{T}_2=2\pi \sqrt{\frac{4\text{ kg}}{\text{0}\text{.15 kN/m}}}\\ =2\pi \sqrt{\frac{4\text{ kg}}{\left(\text{0}\text{.15 kN/m}\right)\left(\frac{1000\text{ N}}{1\text{kN}}\right)}}\\ =2\pi \sqrt{\frac{4}{150}}\\ =1.0\text{ s}\end{array}$

Conclusion:

The time period of vibration is $1.0\text{ s}$.

To determine

The maximum speed and maximum acceleration of the vibrating object when the mass of $1.5\text{kg}$ causing a stretch of $10\text{ cm}$ is replaced by the mass of $4\text{kg}$ and is set into a vibration with an amplitude of $12\text{ cm}$.

Answer to Problem 32SP

Solution: $0.73\text{ m/s}$, $4.5{\text{ m/s}}^2$

Explanation of Solution

Given data:

The mass of $1.5\text{kg}$ is attached to the spring and causes a stretch of $10\text{ cm}$.

The $1.5\text{kg}$ mass is replaced by the $4\text{kg}$ mass and is set into a vibration with an amplitude of $12\text{ cm}$.

Formula used:

The expression for acceleration of a mass in a spring mass system is written as,

$a=-\left(\frac{k}{m}\right)x$

Here, $x$ is the displacement of the mass from the mean position, $k$ is the spring constant, $m$ is the mass of the body, and $a$ is the acceleration of the body.

The expression for velocity of mass in SHM at a location $x$ is 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 $x$, and $x_0$ is the amplitude or the maximum displacement of the mass from the mean position.

Explanation:

Consider the expression for maximum acceleration of the mass $m_2$

$a_2=-\left(\frac{k}{m_2}\right)x_2$

Here, $a_2$ is the maximum accelerationfor vibration for the mass $m_2$.

Substitute $4\text{ kg}$ for $m_2$, $12\text{ cm}$ for $x_2$, and $\text{0}\text{.15 kN/m}$ for $k$

$\begin{array}{c}{a}_2=-\left(\frac{\text{0}\text{.15 kN/m}}{4\text{ kg}}\right)\left(12\text{ cm}\right)\\ =-\left(\frac{\text{0}\text{.15 kN/m}}{4\text{ kg}}\right)\left(\frac{1000\text{ N}}{1\text{ kN}}\right)\left(12\text{ cm}\right)\left(\frac{0.01\text{ m}}{1\text{ cm}}\right)\\ =-\left(\frac{\text{150 N/m}}{4\text{ kg}}\right)\left(0.12\text{ m}\right)\\ =-4.5{\text{ m/s}}^2\end{array}$

The negative sign shows that the acceleration is in the direction opposite to elongation. Therefore, the maximum acceleration is $4.5{\text{ m/s}}^2$.

Consider the expression for velocity of the mass $m_2$

$v_2=\sqrt{\left(x_2^2-x^2\right)\left(\frac{k}{m_2}\right)}$

Here, $v_2$ is the velocity of the mass $m_2$ at a distance $x$ from the equilibrium position.

Understand that the velocity is maximum at equilibrium position. Therefore, substitute 0 for $x$, $4\text{ kg}$ for $m_2$, $12\text{ cm}$ for $x_2$, and $\text{0}\text{.15 kN/m}$ for $k$.

$\begin{array}{c}{v}_2=\sqrt{\left[{\left(12\text{ cm}\right)}^2-{\left(0\right)}^2\right]\left(\frac{\text{0}\text{.15 kN/m}}{4\text{ kg}}\right)}\\ =\sqrt{\left[144{\text{ cm}}^2-0\right]\left(\frac{\text{0}\text{.15 kN/m}}{4\text{ kg}}\right)\left(\frac{1000\text{ N}}{1\text{ kN}}\right)}\\ =\sqrt{\left[144{\text{ cm}}^2\right]\left(\frac{{10}^{-4}{\text{ m}}^2}{1{\text{ cm}}^2}\right)\left(\frac{\text{0}\text{.15 kN/m}}{4\text{ kg}}\right)\left(\frac{1000\text{ N}}{1\text{ kN}}\right)}\\ =0.73\text{ m/s}\end{array}$

Conclusion:

The maximum acceleration is $4.5{\text{ m/s}}^2$ and the maximum velocity is $0.73\text{ m/s}$.

To determine

The speed and acceleration of the vibrating object when displacement is $9\text{ cm}$ considering the mass of $1.5\text{kg}$, stretched by $10\text{ cm}$ is replaced by the mass of $4\text{kg}$ and is set into a vibration with an amplitude of $12\text{ cm}$.

Answer to Problem 32SP

Solution: The acceleration when displacement is $9\text{ cm}$ is $3.3{\text{ m/s}}^2$ and the velocity when displacement is $9\text{ cm}$ is $0.48\text{ m/s}$.

Explanation of Solution

Given data:

The mass of $1.5\text{kg}$ is attached to the spring and is stretched by $10\text{ cm}$.

The $1.5\text{kg}$ mass is replaced by the $4\text{kg}$ mass and is set into a vibration with an amplitude of $12\text{ cm}$.

Formula used:

The expression for acceleration of a mass in a spring mass system is written as,

$a=-\left(\frac{k}{m}\right)x$

Here, $x$ is the displacement of the mass from the mean position, $k$ is the spring constant, $m$ is the mass of the body, and $a$ is the acceleration of the body.

The expression for velocity of mass in SHM at a location $x$ is 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 $x$ and $x_0$ is the amplitude or the maximum displacement of the mass from the mean position.

Explanation:

Consider the expression for acceleration of the mass $m_2$ at a displacement $x$ from the equilibrium position

$a_1=-\left(\frac{k}{m_2}\right)x$

Here, $a_1$ is the accelerationat a displacement $x$.

Substitute $4\text{ kg}$ for $m_2$, $9\text{ cm}$ for $x$, and $\text{0}\text{.15 kN/m}$ for $k$

$\begin{array}{c}{a}_1=-\left(\frac{\text{0}\text{.15 kN/m}}{4\text{ kg}}\right)\left(9\text{ cm}\right)\\ =-\left(\frac{\text{0}\text{.15 kN/m}}{4\text{ kg}}\right)\left(\frac{1000\text{ N}}{1\text{ kN}}\right)\left(9\text{ cm}\right)\left(\frac{0.01\text{ m}}{1\text{ cm}}\right)\\ =-\left(\frac{\text{150 N/m}}{4\text{ kg}}\right)\left(0.09\text{ m}\right)\\ =-3.3{\text{ m/s}}^2\end{array}$

The negative sign shows that the acceleration is in the direction opposite to elongation. Therefore, the acceleration when displacement is $9\text{ cm}$ is $3.3{\text{ m/s}}^2$.

Consider the expression for velocity of the mass $m_2$.

$v_1=\sqrt{\left(x_2^2-x^2\right)\left(\frac{k}{m_2}\right)}$

Here, $v_1$ is the velocity of the mass $m_2$ at a distance $x$ from the equilibrium position.

Substitute $9\text{ cm}$ for $x$, $4\text{ kg}$ for $m_2$, $12\text{ cm}$ for $x_2$, and $\text{0}\text{.15 kN/m}$ for $k$.

$\begin{array}{c}{v}_1=\sqrt{\left[{\left(12\text{ cm}\right)}^2-{\left(9\text{ cm}\right)}^2\right]\left(\frac{\text{0}\text{.15 kN/m}}{4\text{ kg}}\right)}\\ =\sqrt{\left[144{\text{ cm}}^2-81{\text{ cm}}^2\right]\left(\frac{\text{0}\text{.15 kN/m}}{4\text{ kg}}\right)\left(\frac{1000\text{ N}}{1\text{ kN}}\right)}\\ =\sqrt{\left[63{\text{ cm}}^2\right]\left(\frac{{10}^{-4}{\text{ m}}^2}{1{\text{ cm}}^2}\right)\left(\frac{\text{0}\text{.15 kN/m}}{4\text{ kg}}\right)\left(\frac{1000\text{ N}}{1\text{ kN}}\right)}\\ =0.48\text{ m/s}\end{array}$

The velocity when displacement is $9\text{ cm}$ is $0.48\text{ m/s}$.

Conclusion:

The acceleration when displacement is $9\text{ cm}$ is $3.3{\text{ m/s}}^2$ and the velocity when the displacementis $9\text{ cm}$ is $0.48\text{ m/s}$.