Exams › NEET › Physics

A body of mass M executes vertical SHM with periods t₁ and t₂, when separately attached to spring A and spring B respectively. The period of SHM, when the body executes SHM, as shown in the figure is t₀. Then:

1. t₀⁻² = t₁⁻² + t₂⁻²
2. t₀ = t₁ + t₂
3. t₀² = t₁² + t₂²
4. t₀² = t₁⁻² + t₂⁻²

Correct answer: t₀⁻² = t₁⁻² + t₂⁻²

Solution

When two springs are connected in parallel, the effective spring constant is the sum of the individual spring constants. The angular frequency of SHM is proportional to the square root of the spring constant. Using the relationship between time period and angular frequency, we derive that t₀⁻² = t₁⁻² + t₂⁻².

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 →