Question

1. A 50-cm-long spring is suspended from the ceiling. A 230g mass is connected to the end and held at rest with the spring unstretched. The mass is released and falls, stretching the spring by 18cm before coming to rest at its lowest point. It then continues to oscillate vertically.

a. What is the spring constant? (K=)

b. What is the amplitude of the oscillation?

c. What is the frequency of the oscillation?

2. Suppose the free-fall accelaration at some location on earth was exactly 9.8000 m/s2. What would it be at the top of a 1100-m-tall toawer at this location?

 

Answer 1

Concept and reason

The concepts required to solve the given problem is conservation of energy, potential energy of stretched spring, gravitational potential energy, frequency of oscillation in a spring attached to a mass, and variation of acceleration due to gravity with altitude.

Initially, equate variation in gravitational potential energy on mass attached to spring and the elastic potential energy of stretched spring by using conservation of energy. Then, substitute the given values in the equation and solve for spring constant. After that, find the amplitude by taking half of the value of stretching of spring from equilibrium and frequency of oscillation by using expression for frequency of oscillation in a stretched string in terms of mass and spring constant. Finally, find the acceleration due to gravity at a given altitude by using expression for variation of acceleration due to gravity with altitude.

Fundamental

The total mechanical energy of a physical system is conserved when there is no an external force like dissipative forces or frictional forces acting on the system.

The variation in gravitational potential energy U on an object of mass m with change in height $\Delta h$ is given as follows:

$U = mg\Delta h$

Here, g is the acceleration due to gravity.

The potential energy ${U_s}$ of a spring stretched by a distance x from equilibrium position is proportional to square of the stretching x in the spring.

${U_s} = \frac{1}{2}k{x^2}$

Here, k is the spring constant.

The frequency of oscillation in a spring attached to a mass is depends on mass attached to the spring and spring constant. The frequency f is proportional to square root of the spring constant k and inversely proportional to square root of the mass m attached to the spring.

$f = \frac{1}{{2\pi }}\sqrt {\frac{k}{m}}$

The acceleration due to gravity decreases as move away from the earth. The variation of acceleration due to gravity ${g_h}$ with altitude h above the surface of the Earth is given as follows:

${g_h} = \frac{{GM}}{{{{\left( {R + h} \right)}^2}}}$

Here, G is the gravitational constant, M is the mass of the earth, and R is the radius of the earth.

(1)

Using conservation of energy, the variation in gravitational potential energy U of attached mass is equal to potential energy ${U_s}$ of the spring.

$U = {U_s}$

Assume that the change in position of spring from vertical equilibrium position is $\Delta y.$

The variation in gravitational potential energy U of attached mass due to vertical displacement $\Delta y$ is given as follows:

$U = mg\Delta y$

The potential energy ${U_s}$ of spring due to vertical displacement $\Delta y$ from equilibrium position is given as follows:

${U_s} = \frac{1}{2}k{\left( {\Delta y} \right)^2}$

Substitute $mg\Delta y$ for U, and $\frac{1}{2}k{\left( {\Delta y} \right)^2}$ for ${U_s}$ in the above equation and solve for k.

$\begin{array}{c}\\mg\Delta y = \frac{1}{2}k{\left( {\Delta y} \right)^2}\\\\k = \frac{{2mg}}{{\Delta y}}\\\end{array}$

Substitute 230 g for m, $9.8\,{\rm{m/}}{{\rm{s}}^2}$ for g, and 18 cm for $\Delta y$ in the above equation and solve for k.

$\begin{array}{c}\\k = \frac{{2\left( {230\,{\rm{g}}} \right)\left( {\frac{{1\,{\rm{kg}}}}{{1000\,{\rm{g}}}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)}}{{18\,{\rm{cm}}\left( {\frac{{1\,{\rm{m}}}}{{100\,{\rm{cm}}}}} \right)}}\\\\ = 25\,{\rm{N/m}}\\\end{array}$

The amplitude A of oscillation is equal to half of the vertical displacement $\Delta y$ from equilibrium before coming to rest.

$A = \frac{{\Delta y}}{2}$

Substitute 18 cm for $\Delta y$ in the above equation and solve for A.

$\begin{array}{c}\\A = \frac{{18\,{\rm{cm}}\left( {\frac{{1\,{\rm{m}}}}{{100\,{\rm{cm}}}}} \right)}}{2}\\\\ = 0.09\,{\rm{m}}\\\end{array}$

The expression for frequency f of oscillation in terms of mass m and spring constant k is given as follows:

$f = \frac{1}{{2\pi }}\sqrt {\frac{k}{m}}$

Substitute 25.0 N/m for k, and 230 g for m in the above equation and solve for f.

$\begin{array}{c}\\f = \frac{1}{{2\pi }}\sqrt {\frac{{25.0\,{\rm{N/m}}}}{{230\,{\rm{g}}\left( {\frac{{1\,{\rm{kg}}}}{{1000\,{\rm{g}}}}} \right)}}} \\\\ = 1.66\,{\rm{Hz}}\\\end{array}$

(2)

The variation of acceleration due to gravity g with altitude h is given as follows:

${g_h} = \frac{{GM}}{{{{\left( {R + h} \right)}^2}}}$

Substitute $6.6726 \times {10^{ - 11}}\,{\rm{N}} \cdot {{\rm{m}}^2}/{\rm{k}}{{\rm{g}}^2}$ for G, $5.9742 \times {10^{24}}\,{\rm{kg}}$ for M, $6.3781 \times {10^6}\,{\rm{m}}$ for R, and 1100 m for h in the above equation and solve for ${g_h}.$

$\begin{array}{c}\\{g_h} = \frac{{\left( {6.6726 \times {{10}^{ - 11}}\,{\rm{N}} \cdot {{\rm{m}}^2}/{\rm{k}}{{\rm{g}}^2}} \right)\left( {5.9742 \times {{10}^{24}}\,{\rm{kg}}} \right)}}{{{{\left( {6.3781 \times {{10}^6}\,{\rm{m}} + 1100\,{\rm{m}}} \right)}^2}}}\\\\ = 9.7958\,{\rm{m/}}{{\rm{s}}^2}\\\end{array}$

Ans: Part 1.a

The value of spring constant is 25 N/m.

Part 1.b

The amplitude of oscillation is 0.09 m.

Part 1.c

The frequency of oscillation is 1.66 Hz.

Part 2

The acceleration due to gravity at top of the tower is $9.7958\,{\rm{m/}}{{\rm{s}}^2}.$