Question

5. [Honors] A 0.5 kg mass on a spring has velocity as a function of time given by

Answer 1

Given:

The mass, m=0.5 kg

The velocity is given by,

`v_x(t)=-(3.6\text{ cm/s})\sin \lbrack(4.71\text{ rad/s})t-(\pi)/(2)\rbrack\text{ }\rightarrow\text{ (i)}`

To find:

A) Period

B) Amplitude

C) Maximum acceleration of the mass.

D) Force constant of the spring.

Step-by-step explanation:

The displacement of a mass in simple harmonic motion is given by the equation,

`x=A\cos \lbrack\omega t-\phi\rbrack`

Where A is the amplitude, ω is the angular velocity, t is the time, and φ is the phase difference.

By differentiating the above equation, we get the velocity of the object in simple harmonic motion.

Thus,

`v_x(t)=-A\omega\sin \lbrack\omega t-\phi\rbrack\text{ }\rightarrow\text{ (ii)}`

On comparing equation (i) and equation (ii),

`\begin{gathered} A\omega=3.6\text{ cm/s} \\ \omega=4.7\text{ rad/s} \\ \phi=(\pi)/(2) \end{gathered}`

A)

The period is given by,

`T=(2\pi)/(\omega)`

On substituting the known values,

`\begin{gathered} T=(2\pi)/(4.7) \\ =1.34\text{ s} \end{gathered}`

B)

We know,

`A\omega=3.6\text{ cm/s}`

On substituting the known values,

`\begin{gathered} A*4.7=3.6 \\ \Rightarrow A=(3.6)/(4.7) \\ =0.77\text{ cm} \end{gathered}`

C)

The maximum acceleration of a mass in SHM is given by,

`a_(\max )=A\omega^2`

On substituting the known values,

`\begin{gathered} a_(\max )=0.77*10^(-2)*4.7^2 \\ =0.17\text{ m/s}^2 \end{gathered}`

D)

The period of the spring is given by,

`T=2\pi\sqrt[]{(m)/(k)}`

Where k is the force constant of the spring.

On rearranging the above equation,

`\begin{gathered} (T^2)/(4\pi^2)=(m)/(k) \\ \Rightarrow k=(m*4\pi^2)/(T^2) \end{gathered}`

On substituting the known values,

`\begin{gathered} k=(0.5*4\pi^2)/(1.34^2) \\ =11\text{ N/m} \end{gathered}`

Final answer:

A) The period is 1.34 s

B) The amplitude is 0.77 cm

C) The maximum acceleration of the mass is 0.17 m/s²

D) The force constant is 11 N/m