Question

13.-3 points SerCP10 13.P031. МУ Notes Ask Your T A 1.84-kg object on a frictionless from its equilibrium position and then released horizontal track is attached to the end of a horizontal spring whose force constant is 5.00 N/m. The object is displaced 3.52 m to the right attached to initiating simple harmonic motion. (a) What is the force (magnitude and direction) acting on the object 3.50 s after it is released? magnitude direction Direction (b) How many times does the object oscillate in 3.50 s? (Do not round your answer to an integer.) Need Help? Pesth 14. -2 points SerCP10 13 P035 My Notes Ask Your Teack A simple pendulum has a length of 52.3 cm and makes 82.1 complete oscillations in 2.00 min. (a) Find the period of the pendulum. (b) Find the value of g at the location of the pendulum. m/s? Need Help? oMS

Please help with any or all problems I can’t seem to figure them out thanks

Answer 1

for question 14.

as it completes 82.1 oscillation in 120 seconds.

as we know frequency is defined as cycle per second, so we can say f= 82.1/120 = 0.6842 per second

now time period = 1/f

so period of the pendulum= 1/.6842 = 1.462

now time period is given by

$T=2\pi \sqrt{\frac{L}{g}}$

here we know L and T and we need to find g at that particular place. hence put the values.

$1.462=2\pi \sqrt{\frac{52.3*10^{-2}}{g}}\Rightarrow \frac{1.462}{2\pi }=\sqrt{\frac{.523}{g}}$

$0.232=\sqrt{\frac{.523}{g}}\Rightarrow 0.0538=\frac{0.538}{g}$

$g=\frac{.538}{.0523}=10.28$

so value of g at that particular place will be 10.28.

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