Question

At t = 0, an 885-g mass at rest on the end of a horizontal spring (k = 184 n/m) is struck by a hammer which gives it an initial speed of 2.26 m/s. Determine (a) the period and frequency of the motion, (b) the amplitude, (c) the maximum acceleration, (d) the total energy, and (e) the kinetic energy when x = 0.40a where a is the amplitude.

Answer 1

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Answer 2

Given:

Mass of the object $=885g$

Spring constant $K=184N/m$

Velocity of the spring $=2.26m/s$

To find:

1. Time period and frequency of the motion of the spring
2. Amplitude of the spring.
3. Maximum acceleration of the spring.
4. Total energy of the spring.
5. Kinetic energy of the spring.

Solution:

Step 1

We have been given that a mass of $885g$ is attached to the end of the spring of spring constant $184N/m$.

We know, time period of a spring is given by

$T=$  $2$ π$\sqrt{\frac{M}{K} }$

We have,

$M=885g=0.885kg$

$K=184N/m$

Hence,

$T=2(3.14)\sqrt{\frac{0.885}{184} }$

$T=6.28$ × $0.07$

$T=0.44s$  $(approx)$

Hence, the time period of the spring is 0.44 seconds.

Frequency is given as the reciprocal of the time period.

Hence,

$Frequency=\frac{1}{0.44}$

$Frequency =2.28Hz$

Hence, the frequency of the spring is 2.28 oscillations per second.

Explanation 2

We have been given that when the mass is at rest at the mean position, it is truck by the hammer, the spring moves with a velocity of 2.26 m/s.

We know, velocity of oscillation of the wave is given by

$v=$ ω$\sqrt{A^{2}- x^{2} }$

Here, we have,

$v=2.26m/s$      $;$ $x=0$       $;$

ω $=2$ π$f$ $=2(3.14)(2.28)=14.3rad/s$

We get

$2.26=14.3\sqrt{A^{2}- (0)^{2} }$

$A=\frac{2.26}{14.3}$

$A=0.158m$

$A=15.8cm$

Explanation 3

The acceleration of a wave is directly proportional to the displacement of the particle from its mean position.

$a$ ∝ $x$

$a=-$ω²$x$

Here, according to the question, $x$ is the maximum displacement or the amplitude of the spring or $x=0.158$   and  ω$=14.3rad/s$

$a=-(14.3)^{2}(0.158)$

$a=-(204.5)(0.158)$

$a=32.311m/s^{2}$

Hence, the maximum acceleration of the spring will be 32.311 m/s².

Explanation 4

Total energy of the spring is the energy stored in the spring on getting compressed or stretched by any external force. It is given by

$E=\frac{1}{2}kx^{2}$

where, $k$ is the spring constant.

$E=\frac{1}{2}(184)(0.158)^{2}$

$E=2.296J$

Step 5

We have been given the position of the mass that is $x=0.40A$ where, $A$ is the amplitude of the string.

We already know, velocity is given by

$v=$ ω$\sqrt{A^{2}- x^{2} }$

Substituting the known values in the equation, we get

$v=(14.3)\sqrt{A^{2}- (0.4A)^{2} }$

$v^{2} =(14.3)0.84A^{2}$

Now,

The kinetic energy is given as

$KE=\frac{1}{2} mv^{2}$

$KE=\frac{1}{2}m(14.3)(0.84A^{2} )$

$KE=\frac{1}{2}(0.885)(14.3)(0.84)(0.158)^{2}$

$KE=\frac{1}{2}(0.26538)$

$KE=0.13269J$

Hence, the kinetic energy of the spring when x = 0.4 A is 0.132 J.

Final answer:

Hence,

1. The time period and frequency of the spring are 0.44 s and 2.28 Hz respectively.
2. The amplitude of the wave motion is 0.158 m.
3. The maximum acceleration of the spring is 32.311 m/s².
4. The total energy of the spring is 2.29 J.
5. The kinetic energy of the spring at x = 0.4 A is 0.132 J.