Respuesta :
The car’s frequency oscillation would be 2√ times larger if the springs are replaced with stiffer ones and this would double the spring constant. Period and frequency of a simple harmonic oscillator are independent of amplitude. The period is always related to how stiff the system is, the more stiff the object is, the smaller the period and the larger the oscillation frequency.
The car’s oscillation frequency would be [tex]\sqrt{2}[/tex] times more if stiffer string is used.
Further Explanation:
Simple Harmonic Motion: Simple Harmonic Motion or SHM is a periodic motion in which restoring force always tended towards the mean position or equilibrium position. The periodic motion is that motion which repeats its path after a fixed interval of time. The example of periodic motion is motion of the hands of a clock.
The formula to calculate the angular frequency of the spring with spring constant [tex]k[/tex] and [tex]m[/tex] is given as,
[tex]\boxed{w=\sqrt{\frac{k}{m}}}[/tex]
The angular frequency of the suspension spring is calculated as follows:
[tex]\begin{aligned}w_{\text{suspension}}&=\sqrt{\frac{k_{\text{suspension}}}{m}}\\&=\sqrt{\frac{k}{m}\end{aligned}[/tex]
The angular frequency of the stiffer spring is calculated as follows:
[tex]\begin{aligned}w_{\text{stiffer}}&=\sqrt{\frac{k_{\text{stiffer}}}{m}}\\&=\sqrt{\frac{2k}{m}\end{aligned}[/tex]
Therefore, the car’s oscillation frequency would be [tex]\sqrt{2}[/tex] times more if stiffer string is used.
Learn More:
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Answer Details:
Grade: High School
Subject: Physics
Chapter: Simple Harmonic Motion
Keywords:
Simple harmonic motion, suspension spring, stiffer, angular frequency, motion, force, time period, spring constant, doubled, car, mass, m, bump, oscillates, oscillation frequency, speed bump