Linear SHM

A spring balance has a scale that reads from $0$ to $50 \mathrm{kg}$. The length of the scale is $20 \mathrm{cm}$. A body suspended from this spring, when displaced and released, oscillates with a period of $0.6 \mathrm{s}$. What is the weight of the body?

A body of mass $m$ is attached to the lower end of a spring whose upper end is fixed. The spring has negligible mass. When the mass $m$ is slightly pulled down and released, it oscillates with a time period of $3 \mathrm{s}$. When the mass $m$ is increased by $1 \mathrm{kg}\text{,}$ the time period of oscillations becomes $5 \mathrm{s}$. The value of $m \mathrm{in} \mathrm{kg}$ is

A horizontal spring is connected to a mass $M$. It executes simple harmonic motion. When the mass $M$ passes through its mean position, an object of mass $m$ is put on it and the two move together. The ratio of frequencies before and after will be:

A spring is oscillating with frequency $4 \mathrm{Hz}$ having spring constant $k$. An identical spring is connected in series in new system as shown in figure. Find the new frequency.

When mass m oscillates with spring of spring constant k₁, its frequent is 3Hz and with spring of spring constant k₂ its frequency is 6Hz. Then frequency of oscillations when mass is connected with springs as shown in figure is :-

Two similar sprirgs P and Q have spring constants K_P and K_Q, such that K_P > K_Q . The are stretched, first by the same amount (case a,) then by the same force (case b). The work done by the springs W_P and W_Q are related as, in case (a) and case (b), respectively: 

A mass of $10g$ is connected to a massless spring then time period of small oscillation is $10 \mathrm{second}$. If$10 g$ mass is replaced by $40 g$ mass in same spring, then its time period Ml be :-

The spring constant of two springs are K₁ and K₂ respectively springs are stretch up to that limit when potential energy of both becomes equal. The ratio of applied force (F₁ and F₂) on them will be :

Mass $m$ is suspended from a spring of force constant $K$. Spring is cut into two equal parts end same mass is suspended from it, her new frequency will be:

On loading a spring with bob, its period of oscillation in a vertical plane is $T$. If this spring pendulum is tied with one end to a friction less table and made to oscillate in a horizontal plane, its period of oscillation will be :

The time period of a spring pendulum on earth is $T$. If it is taken on the moon, and made to oscillate, the period of vibration will be :

A spring is made to oscillate after suspending a mass $m$ from one of its ends. The time period obtained is $2$ seconds. On increasing the mass by $2\mathrm{kg}$, the period of oscillation is increased by$1\mathrm{s}$. The initial mass $m$ will be :

In the adjoining figure, the frequency of oscillation for a mass $M$ will be proportional to 

On suspending a mass $M$ from a spring of force constant $K$, frequent of vibration $f$ is obtained. If a second spring as shown in the figure, is arranged then the frequency will be :

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A block of mass 'm' moving with velocity (v) collides perfectly inelastically with another identical block attached to spring of force constant K. What will be the amplitude of resulting SHM?

When a mass of 1 kg is suspended from a vertical spring, its length increases by 0.98 m. If this mass is pulled downwards and then released, what will be periodic time of vibration of spring? (g = 9.8 m/s²)

The period of oscillation of a mass $M$ suspended from a spring of negligible mass is $T$. If along with it another mass $M$ is also suspended, the period of oscillation will now be

Two particles $A$ and $B$ of equal masses are suspended from two massless springs of spring constants $k_1$ and $k_2$, respectively. If the maximum velocities, during oscillations are equal, the ratio of amplitudes of $A$ and $B$ is