Exams › NEET › Physics

A mass m is vertically suspended from a spring of negligible mass; the system oscillates with a frequency n. What will be the frequency of the system, if a mass 4m is suspended from the same spring?

1. n
2. 4n
3. n/2
4. 2n

Correct answer: n/2

Solution

The frequency of oscillation for a spring-mass system is given by \( n = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \). When the mass is increased to 4m, the new frequency becomes \( n' = \frac{1}{2\pi} \sqrt{\frac{k}{4m}} = \frac{n}{2} \).

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 →