PHYSICS COLLEGE

A block-spring system consists of a spring with constant k = 475 N/m attached to a 2.50 kg block on a frictionless surface. The block is pulled 5.50 cm from equilibrium and released from rest. For the resulting oscillation, find the amplitude, angular frequency, frequency, and period. What is the maximum value of the block's velocity and acceleration?

Answers

Answer 1

Answer:

Explanation:

It is given that,

Spring constant of the spring, k = 475 N/m

Mass of the block, m = 2.5 kg

Elongation in the spring from equilibrium, x = 5.5 cm

(a) We know that the maximum elongation in the spring is called its amplitude. So, the amplitude for the resulting oscillation is 5.5 cm.

(b) Let $\omega$ is the angular frequency. It is given by :

$\omega=\sqrt{(k)/(m)}$

$\omega=\sqrt{(475)/(2.5)}$

$\omega=13.78\ rad/s$

(c) Let T is the period. It is given by :

$T=(2\pi)/(\omega)$

$T=(2\pi)/(13.78)$

T = 0.45 s

(d) Frequency,

$f=(1)/(T)$

$f=(1)/(0.45)$

f = 2.23 Hz

(e) Let v is the maximum value of the block's velocity. It is given by :

$v_(max)=\omega* A$

$v_(max)=13.78* 5.5* 10^(-2)$

$v_(max)=0.757\ m/s$

The value of acceleration is given by :

$a=\omega^2A$

$a=(13.78)^2* 5.5* 10^(-2)$

$a=10.44\ m/s^2$

Hence, this is the required solution.

Answer 2

Answer:

Final answer:

The amplitude of the block-spring system is 0.055 m, with an angular frequency of 13.77 rad/s, and a frequency of approximately 2.19 Hz. The system has a period of approximately 0.46 s, with the maximum velocity being 0.76 m/s, and the maximum acceleration being 189 m/s².

Explanation:

In this block-spring system, we can determine the oscillation properties with the given parameters: mass (m = 2.50 kg), spring constant (k = 475 N/m), and displacement (x = 5.50 cm = 0.055 m).

Amplitude (A): This is the maximum displacement from the equilibrium position, in this case, it is 0.055 m, the distance from which the block is pulled and released.
Angular Frequency (ω): This can be calculated by the formula ω = √(k/m) which is equal to √(475/2.50) = 13.77 rad/s.
Frequency (f): It is given by the formula f = ω / 2π ≈ 2.19 Hz.
Period (T): The time for one complete cycle of the motion, calculated as T = 1/f ≈ 0.46 s.
Maximum value of velocity (v max): Calculated by the formula v = ω*A = 13.77 * 0.055 = 0.76 m/s.
Maximum value of acceleration (a max): Maximum acceleration occurs when the block is at the maximum displacement (amplitude), found with formula a = ω²*A = 189 m/s².

Learn more about Oscillations here:

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Answers

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