In fig.15-28, a spring–block system is put into SHM in two experiments. In the first, the block is pulled from the equilibrium position through a displacement and then released. In the second, it is pulled from the equilibrium position through a greater displacementd2 and then released. Are the (a) amplitude, (b) period, (c) frequency, (d) maximum kinetic energy, and (e) maximum potential energy in the second experiment greater than, less than, or the same as those in the first experiment?

Short Answer

Expert verified

a) The amplitude of second experiment is greater than the first experiment.

b) The period for both the experiments is same.

c) The frequency for both the experiments is same.

d) The maximum kinetic energy of second experiment is greater than the first experiment.

e) The maximum potential energy of second experiment is greater than the first experiment

Step by step solution

01

The given data 

The figure for the block-spring system is given.

02

Understanding the concept of equations of a SHM

We can use the formulae for amplitude, period, frequency, kinetic energy, and potential energy related to displacement. Comparing the displacements of the system in both experiments, we can compare the above quantities.

Formulae:

The displacement equation of a body in SHM,x=xmcosωt+ϕ (i)

The acceleration of the body in SHM, SHM,a=-ω2x (ii)

The period of oscillation in SHM, SHM,T=2πmk (iii)

The frequency of the oscillation in SHM, SHM,f=1T (iv)

The maximum kinetic energy of a system, K.Emax=12kxm2 (v)

The maximum potential energy of a system, Umax=12kxm2 (vi)

03

Calculation of the amplitudes of the experiments

a)

Here, displacements are d1andd2 Now, from equation (i), we can get that

xαxm

Since d2>d1, the above proportionality equation of both the experiments can be given as:

xm2>xm1

Hence, the amplitude of second experiment is greater than the first experiment.

04

Calculation of the period of the experiments 

b)

From equation (iii), we can see that m and k are same for both systems, so period will not change, that is given by,T1=T2

Hence, the period remains the same in both experiments.

05

Calculation of the frequency of the experiments 

c)

From equation (iv), we can see that the period value is inversely proportional to frequency.

Again, as the period is same in both experiments, the frequency will also be the same.

06

Calculation of the maximum kinetic energy of the experiments 

d)

Kinetic energy is directly proportional to displacement from equation (v) and displacement is proportional to amplitude from part (a) calculations.

We have,d2>d1.Thus, we get the kinetic energies as:

K.Emax2>K.Emax1

Hence, the maximum kinetic energy of second experiment is greater than the first experiment.

07

Calculation of the maximum potential energy of the experiments 

e)

Potential energy is also directly proportional to displacement from equation (vi)

We have,d2>d1.Therefore, we get the potential energies as:

Umax2>Umax1

Hence, the maximum potential energy of second experiment is greater than the first experiment.

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Most popular questions from this chapter

A 0.10 kgblock oscillates back and forth along a straight line on a frictionless horizontal surface. Its displacement from the origin is given byx=(10cm)cos[(10rad/s)t+ττ/2rad]. (a) What is the oscillation frequency? (b) What is the maximum speed acquired by the block? (c) At what value of x does this occur? (d) What is the magnitude of the maximum acceleration of the block? (e) At what value of x does this occur? (f) What force, applied to the block by the spring, results in the given oscillation?

Figure 15-29gives, for three situations, the displacements of a pair of simple harmonic oscillators (A and B) that are identical except for phase. For each pair, what phase shift (in radians and in degrees) is needed to shift the curve for A to coincide with the curve for B? Of the many possible answers, choose the shift with the smallest absolute magnitude.

The vibration frequencies of atoms in solids at normal temperatures are of the order of1013Hz. Imagine the atoms to be connected to one another by springs. Suppose that a single silver atom in a solid vibrates with this frequency and that all the other atoms are at rest. Compute the effective spring constant. One mole of silver (6.021023atoms) has a mass of 108 g.

A torsion pendulum consists of a metal disk with a wire running through its center and soldered in place. The wire is mounted vertically on clamps and pulled taut. 15-58a Figuregives the magnitude τof the torque needed to rotate the disk about its center (and thus twist the wire) versus the rotation angle θ. The vertical axis scale is set by τs=4.0×10-3N.m.=.The disk is rotated to θ=0.200rad and then released. Figure 15-58bshows the resulting oscillation in terms of angular position θversus time t. The horizontal axis scale is set by ts=0.40s. (a) What is the rotational inertia of the disk about its center? (b) What is the maximum angular speedof dθ/dtthe disk? (Caution: Do not confuse the (constant) angular frequency of the SHM with the (varying) angular speed of the rotating disk, even though they usually have the same symbol. Hint: The potential energy U of a torsion pendulum is equal to 12kθ2, analogous to U=12kx2for a spring.)

Which of the following relationships between the acceleration a and the displacement x of a particle involve SHM: (a) a=0.5x, (b) a=400x2, (c) a=20x, (d)a=-3x2?

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