Exams › NEET › Physics

A mass m is suspended from a two coupled springs, connected in series. The force constant for springs are k₁ and k₂. The time period of the suspended mass will be

1. T = 2π√(m / (k₁ − k₂))
2. T = 2π√(mk₁k₂ / (k₁ + k₂))
3. T = 2π√(m / (k₁ + k₂))
4. T = 2π√(m(k₁ + k₂) / k₁k₂)

Correct answer: T = 2π√(mk₁k₂ / (k₁ + k₂))

Solution

When two springs are connected in series, the effective spring constant is given by 1/k_eff = 1/k₁ + 1/k₂, which simplifies to k_eff = (k₁k₂) / (k₁ + k₂). The time period of oscillation is T = 2π√(m / k_eff). Substituting k_eff, we get T = 2π√(mk₁k₂ / (k₁ + k₂)).

Related NEET Physics questions

• One oscillation completed by a vibrating body in one second is known as:
• A body executes simple harmonic motion. The potential energy (P.E.), the kinetic energy (K.E.) and total energy (T.E.) are measured as a function of time t. Which of the following statements is true. Assume that the particle started at \( t=0 \) at its mean position
• The amplitude of a particle executing SHM about O is 10 cm. Then: This question has multiple correct options
• The displacement (in \( \mathrm{m} \) ) of a particle of mass 100 gram from its equilibrium position is given by \( \boldsymbol{y}=\mathbf{0 . 0 1} \sin 2 \pi(t+ \) 0.4). Select the correct option(s): This question has multiple correct options
• Three simple harmonic motions in the same direction having the same amplitude \( A \) and same period are superposed. If each differs in phase from the next by \( 45^{0}, \) then
• A particle executes \( \boldsymbol{S} \boldsymbol{H} \boldsymbol{M} \) of amplitude \( 25 c m \) and time period \( 3 s \) What is the minimum time required for the particle to move between two points \( 12.5 \mathrm{cm} \) on either side of the mean position?

⚔️ Practice NEET Physics free + battle 1v1 →