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A 1.00 kg object is attached to a horizontal spring. the spring is initially stretched by 0.500 m, and the object is released from rest there. it proceeds to move without friction. the next time the speed of the object is zero is 0.100 s later. what is the maximum speed of the object?

Respuesta :

skyluke89
The  spring is initially stretched, and the mass released from rest (v=0). The next time the speed becomes zero again is when the spring is fully compressed, and the mass is on the opposite side of the spring with respect to its equilibrium position, after a time t=0.100 s. This corresponds to half oscillation of the system. Therefore, the period of a full oscillation of the system is
[tex]T=2 t = 2 \cdot 0.100 s = 0.200 s[/tex]
Which means that the frequency is
[tex]f= \frac{1}{T}= \frac{1}{0.200 s}=5 Hz [/tex]
and the angular frequency is
[tex]\omega=2 \pi f = 2 \pi (5 Hz)=31.4 rad/s[/tex]

In a spring-mass system, the maximum velocity of the object is given by
[tex]v_{max} = A \omega[/tex]
where A is the amplitude of the oscillation. In our problem, the amplitude of the motion corresponds to the initial displacement of the object (A=0.500 m), therefore the maximum velocity is
[tex]v_{max} = A \omega = (0.500 m)(31.4 rad/s)= 15.7 m/s[/tex]
snehashish65

The maximum speed of the object is 15.70 m/s

Given data:

The mass of object attached to horizontal spring is, m = 1.00 kg.

The stretching distance is, x = 0.500 m.

Time interval is, t = 0.100 s.

The linear velocity of spring - mass system is given as,

[tex]v = x \times \omega[/tex]

Here, [tex]\omega[/tex] is an angular speed. Solving as,

[tex]v = x \times (2 \pi f )\\\\v = x \times (\dfrac{2 \pi}{T} )[/tex]

Time period (T) for complete oscillation is, [tex]T = 2t[/tex].

[tex]v = x \times (\dfrac{2 \pi}{ 2 t} )\\v = 0.500 \times (\dfrac{ \pi}{0.100} )\\v =15.70 \;\rm m/s[/tex]

Thus, the maximum speed of the object is 15.70 m/s.

Learn more about the oscillation here:

https://brainly.com/question/15780863?referrer=searchResults

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