Question

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

Answer 1

Answer:1080 N/m

Step-by-step 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{(k)/(m)}`

`v_1_(max)=A_1\omega _1`

`\omega _1=\sqrt{(k_1)/(m)}`

similarly
`\omega _2=\sqrt{(k_2)/(m)}`

also
`A_1=2A_2`

so
`A_1* \sqrt{(k_1)/(m)}=A_2* \sqrt{(k_2)/(m)}`

`2* √(k_1)=√(k_2)`

`k_2=4k_1`

`k_2=4* 245=1080 N/m`