This set of Physical Chemistry Multiple Choice Questions & Answers (MCQs) focuses on “Classical Harmonic Oscillator”.

1. For a ball attached on a spring with spring constant k and extended a length x, what is the force F due to the spring?

a) *F = -kx*

b) *F = -kx ^{2}*

c)

*F = \(\frac{-k}{x}\)*

d)

*F = kx*

View Answer

Explanation: The force due to a spring acting on a ball is given by the spring constant multiplied by change in displacement in the opposite direction. There is a negative sign because the direction of displacement and force are opposite to one another. This expression assumes there are no external forces acting on the ball attached.

2. What is the force F on a ball of mass m due to the spring in terms of change in displacement per unit time?

a) *F = ma*

b) *F = m\(\frac{d^2 x}{dt^2}\)*

c) *F = -kx*

d) *F = kx*

View Answer

Explanation: Although

*F = ma*, this is not in terms of change in displacement per unit time. Acceleration is the rate of change of velocity per unit time and velocity is the rate of change of position per unit time; hence acceleration can be written as a double derivative of position per unit time, i.e.

*F = m\(\frac{d^2 x}{dt^2}\)*

3. What is the differential equation force balance for a ball of mass m attached to a spring with spring constant k and felt a change in displacement x?

a) *–\(\frac{k}{m}x\)* = 0

b) *mk\(\frac{d^2 x}{dt^2}\)* = 0

c) *m\(\frac{d^2 x}{dt^2}\) – kx* = 0

d) *m\(\frac{d^2 x}{dt^2}\) + kx* = 0

View Answer

Explanation: Equating the force due to a spring and change in displacement gives

*m\(\frac{d^2 x}{dt^2}\) = -kx;*rearranging this gives

*m\(\frac{d^2 x}{dt^2}\) + kx = 0*

4. What is the general solution to the differential equation for position of the ball attached to a spring as a function of time, spring constant, and angular frequency ω = \(\sqrt{\frac{k}{m}}\)?

a) *x(t) = e ^{-iωt}*

b)

*x(t) = e*

^{iωt}c)

*x(t) = A*sin

*ωt + B*cos

*ωt*

d)

*x(t) = A*tan

*ωt*

View Answer

Explanation: The differential equation m\(\frac{d^2 x}{dt^2}\) + kx = 0 can be solved in various ways. The first is variable separable. It can also be solved by assuming a sin and cosine dependence since the equation only has constants and one x term. Having an exponential term is not incorrect as Euler’s identity can be applied, but it needs to be a sum of two exponential terms such that the imaginary terms cancel.

5. What is the potential energy V of the spring at position x?

a) *V = \(\frac{kx^2}{2}\)*

b) *V = \(\frac{kx^2}{2\pi}\)*

c) *V = -kx*

d) *V = kx*

View Answer

Explanation: Force is the negative derivative of potential energy, i.e.

*F = –\(\frac{dV}{dx}\) —→ V = -∫ Fdx*

Integrating this gives

*V = -[∫ – kxdx] = \(\frac{kx^2}{2}\)*

**Sanfoundry Global Education & Learning Series – Physical Chemistry.**

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