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Yuliya22 [10]
3 years ago
14

Objects of equal mass are oscillating up and down in simple harmonic motion on two different vertical springs. The spring consta

nt of spring 1 is 245 N/m. The motion of the object on spring 1 has twice the amplitude as the motion of the object on spring 2. The magnitude of the maximum velocity is the same in each case. Find the spring constant of spring 2.
Physics
1 answer:
olga55 [171]3 years ago
5 0

Answer:1080 N/m

Explanation:

Given

Spring constant of first Spring K_1=245 N/m

mass of both the system is same

Both system have same maximum velocity

if x=A\sin \omega t is general equation of SHM

then maximum velocity is given by

v_{max}=A\omega

\omega =\sqrt{\frac{k}{m}}

v_1_{max}=A_1\omega _1

\omega _1=\sqrt{\frac{k_1}{m}}

similarly \omega _2=\sqrt{\frac{k_2}{m}}

also A_1=2A_2

so A_1\times \sqrt{\frac{k_1}{m}}=A_2\times \sqrt{\frac{k_2}{m}}

2\times \sqrt{k_1}=\sqrt{k_2}

k_2=4k_1

k_2=4\times 245=1080 N/m

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