A spring mass system is hanging from the ceiling of an elevator in equilibrium. Elevator suddenly starts accelerating upwards with acceleration a. Find frequency, amplitude of the resulting SHM
Answers
Frequency of the resulting SHM is independent of acceleration and is given by,
f = (1/2π)√k/m
where k = spring constant
Initially the spring mass system can be represented by eqn
mg - kx = 0
=> x = mg/k
when the lift starts accelerating upwards with acceleration a, net acceleration will be (g+a), hence the spring mass system can be represented by eqn
m(g+a) - kx₁ = 0
=> x₁ = m(g+a)/k
Hence the amplitude of the SHM will be given as,
A = x₁ - x = m(g+a)/k - mg/k
=> A = ma/k
Hence the frequency of the SHM will be f = (1/2π)√k/m
Amplitude of the SHM will be A = ma/k
Answer:
Frequency of the resulting SHM is independent of acceleration and is given by,
f = (1/2π)√k/m
where k = spring constant
Initially the spring mass system can be represented by eqn
mg - kx = 0
=> x = mg/k
when the lift starts accelerating upwards with acceleration a, net acceleration will be (g+a), hence the spring mass system can be represented by eqn
m(g+a) - kx₁ = 0
=> x₁ = m(g+a)/k
Hence the amplitude of the SHM will be given as,
A = x₁ - x = m(g+a)/k - mg/k
=> A = ma/k
Hence the frequency of the SHM will be f = (1/2π)√k/m
Amplitude of the SHM will be A = ma/k