Question

part h
how much time t does the block take to travel from the point of maximum displacement to the opposite point of maximum displacement?
express your answer in seconds.
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part i
what distance d does the object cover during one period of oscillation?
express your answer in meters.

Explanation:

Step1: Recall properties of simple - harmonic motion

In simple - harmonic motion, the time taken to travel from one extreme (maximum displacement) to the opposite extreme is half of the period ($T$). Let the period be $T$. So, $t=\frac{T}{2}$. But since no value of $T$ is given, if we assume the general case, for a full - cycle (period $T$) which is from one maximum to the next maximum passing through the equilibrium and the other extreme, the time from one extreme to the other is $\frac{T}{2}$.

Step2: Analyze distance in one period

In simple - harmonic motion, if the amplitude of the motion is $A$ (the maximum displacement from the equilibrium position), the object moves from $+A$ to $-A$ and then back to $+A$ in one period. The distance covered from $+A$ to $-A$ is $2A$, and then from $-A$ back to $+A$ is another $2A$. So the total distance $d$ covered in one period is $d = 4A$.

Answer:

For Part H: Without knowing the period $T$, we can only say in terms of the period $t=\frac{T}{2}$
For Part I: Without knowing the amplitude $A$, we can only say $d = 4A$