Question

If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released, it will oscillate. If it is displaced a distance 0.125m from its equilibrium position and released with zero initial speed. Then after a time 0.800s its displacement is found to be a distance 0.125m on the opposite side, and it has passed the equilibrium position once during this interval. Find : Amplitude, period, and frequency.

Answer 1

Explanation:

An object is attached to the spring and then released. It begins to oscillate. If it is displaced a distance 0.125 m from its equilibrium position and released with zero initial speed. The amplitude of a wave is the maximum displacement of the particle. So, its amplitude is 0.125 m.  

After 0.800 seconds, its displacement is found to be a distance 0.125 m on the opposite side. The time period will be, $t=2\times 0.8=1.6\ s$

We know that the relation between the time period and the time period is given by :

$f=\dfrac{1}{t}$

$f=\dfrac{1}{1.6}$

f = 0.625 Hz

So, the frequency of the object is 0.625 Hz. Hence, this is the required solution.

Answer 2

Answer:

The amplitude is 0.125 m, period is 1.64 sec and frequency is 0.61 Hz.

Explanation:

Given that,

Distance = 0.125 m

Time = 0.800 s

Since the object is released form rest, its initial displacement = maximum displacement .

(a). We need to calculate the amplitude

The amplitude is maximum displacement.

$A=0.125 m$

(b). We need to calculate the period

The object will return to its original position after another 0.820 s,

So the time period will be

$T=2\times t$

Put the value into the formula

$T=2\times0.820$

$T=1.64\ sec$

(c). We need to calculate the frequency

Using formula of frequency

$f=\dfrac{1}{T}$

Put the value into the formula

$f=\dfrac{1}{1.64}$

$f=0.61\ Hz$

Hence, The amplitude is 0.125 m, period is 1.64 sec and frequency is 0.61 Hz.