IB Physics SL C.1 Simple harmonic motion questions

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Question 1

Multiple choiceMedium1 mark

An ideal mass–spring system has mass $m$ spring constant $k$ and angular frequency $\omega$ The mass is changed to $m/4$ and the spring constant to $9k$ What is the ratio of the new angular frequency to $\omega$

Show answer and explanation

Answer: D. For a mass–spring system, $T=2\pi\sqrt{m/k}$ and $\omega=2\pi/T$

\frac{T_{\text{new}}}{T} =\sqrt{\frac{(m/4)/(9k)}{m/k}} =\frac{1}{6}, \qquad \frac{\omega_{\text{new}}}{\omega} =\frac{T}{T_{\text{new}}} =6.

A: gives the period ratio but does not take its reciprocal for angular frequency.
B: applies the mass change alone, giving a factor of $2$ and omits the spring change.
C: applies the spring change alone, giving a factor of $3$ and omits the mass change.
Remember: combine both changes when finding the period, then use the inverse relation between angular frequency and period.

Question 2

Paper 2Hard10 marks

A small mass attached to a vertical spring oscillates in simple harmonic motion between upper turning point U and lower turning point L, passing through equilibrium position O. The spring remains stretched throughout the motion. Consider the mass, spring and Earth as the system. Air resistance and other resistive forces are negligible. The mass is released from rest at U. Its mechanical energy consists of kinetic, gravitational potential and elastic potential energy.

(a)

As the mass moves from U to L, state how its gravitational potential energy and elastic potential energy change.

[2]

(b)

At which position, U, O or L, is the kinetic energy greatest?

[1]

(c)

Compare the total potential energy of the system at U and L. Explain your answer with reference to the gravitational and elastic potential energies.

[3]

(d)

A student says: “The gravitational potential energy is lower at L than at U, so mechanical energy has been lost during the motion from U to L.” Evaluate this statement by describing the energy changes over U→O→L→U.

[4]

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