Should the Answer Be B? | Reasoning Explained

[hidemi]

Homework Statement

The angular frequency of a simple pendulum depends on its length and on the local acceleration due to gravity. The rate at which the angular displacement of the pendulum changes, dθ/dt, is:

A. √ (mgL/I)
B. √(g/L)
C. 2兀 (L/g)
D. √(k/m)
E. none of the above

Relevant Equations

F=-kx
T=2兀/w
w = √(g/L)

I think the answer should be B, not E.

Here's the reasoning:
F = -kx = mg sinθ = -mg θ (because θ very small) = -mgx/L
T = 2π*√(m/k) = 2π*√(m/(mg/L)) = 2π*√(L/g)
Also, T= 2π/w and thus, w = dθ/dt = √(g/L)

 

[Steve4Physics]

Here's the reasoning:
F = -kx = mg sinθ = -mg θ (because θ very small) = -mgx/L
T = 2π*√(m/k) = 2π*√(m/(mg/L)) = 2π*√(L/g)
Also, T= 2π/w and thus, w = dθ/dt = √(g/L)

Point 1

For simple harmonic motion (SHM):
The standard equation for the period of a simple pendulum is $T = 2\pi \sqrt \frac L g$.
The standard equation for the period of a mass (m) oscillating on a spring (spring constant k) is $T = 2\pi \sqrt\frac m k$

You seem to be mixing these equations together! By accident, this may give a sensible answer - but it is incorrect to do this.

Point 2

Angular frequency and angular speed are not the same thing, though the same symbol, $\omega$, is often used for both. This causes confusion!

In the context of SHM, angular frequency is given by $\omega = \frac {2\pi} {T}$ where T is the period, Since T is constant for SHM, $\omega$ is constant.

In the context of a changing angle (e.g. circular motion), angular speed is given by $\omega = \frac {d\theta} {dt}$. It is the rate of change of angular displacement. It doesn't have to be constant.

In the context of a pendulum, angular frequency, $\omega$, is not the same thing as $\frac {d\theta} {dt}$

Point 3

Imagine (or even watch!) a pendulum. Do you think $\frac {d\theta} {dt}$ is constant during SHM? Does this help you answer the original question?

 

[hidemi]

Point 1

For simple harmonic motion (SHM):
The standard equation for the period of a simple pendulum is $T = 2\pi \sqrt \frac L g$.
The standard equation for the period of a mass (m) oscillating on a spring (spring constant k) is $T = 2\pi \sqrt\frac m k$

You seem to be mixing these equations together! By accident, this may give a sensible answer - but it is incorrect to do this.

Point 2

Angular frequency and angular speed are not the same thing, though the same symbol, $\omega$, is often used for both. This causes confusion!

In the context of SHM, angular frequency is given by $\omega = \frac {2\pi} {T}$ where T is the period, Since T is constant for SHM, $\omega$ is constant.

In the context of a changing angle (e.g. circular motion), angular speed is given by $\omega = \frac {d\theta} {dt}$. It is the rate of change of angular displacement. It doesn't have to be constant.

In the context of a pendulum, angular frequency, $\omega$, is not the same thing as $\frac {d\theta} {dt}$

Point 3

Imagine (or even watch!) a pendulum. Do you think $\frac {d\theta} {dt}$ is constant during SHM? Does this help you answer the original question?

Thank you. I got it.