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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.

User Ivoputzer
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2 Answers

2 votes

Answer:

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

Step-by-step 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* t

Put the value into the formula


T=2*0.820


T=1.64\ sec

(c). We need to calculate the frequency

Using formula of frequency


f=(1)/(T)

Put the value into the formula


f=(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.

User Russi
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7.7k points
5 votes

Step-by-step 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* 0.8=1.6\ s

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


f=(1)/(t)


f=(1)/(1.6)

f = 0.625 Hz

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

User Amsakanna
by
8.2k points
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