Tag: horizontal oscillations of a mass attached to a spring

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Multiple choice horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

A mass of 10g is connected to a massless spring then time period of small oscillation is 10 second. If 10 g mass is replaced by 40 g mass in same spring then its time period will be :-

  1. 5 s

  2. 10 s

  3. 20 s

  4. 40 s
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The time period of a mass-spring system is T = 2*pi*sqrt(m/k). Since T is proportional to sqrt(m), if the mass increases by a factor of 4 (from 10g to 40g), the time period increases by a factor of sqrt(4) = 2. Thus, 10s becomes 20s.

Multiple choice horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

Two equal masses are connected by a spring satisfying Hooks law and are placed on a frictionless table. The spring is elongated a little and allowed to go. Let the angular frequency of oscillations be $\omega$. Now one of the masses is stopped. The square of the new angular frequency is :

  1. ${\omega}^{2}$
  2. $\dfrac{{\omega}^{2}}{2}$
  3. $\dfrac{{\omega}^{2}}{3}$
  4. $2{\omega}^{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For two masses m connected by a spring k, the effective mass in the oscillation is m/2, so omega^2 = k/(m/2) = 2k/m. If one mass is fixed, the system acts like a single mass m on a spring k, so the new omega'^2 = k/m. Thus, omega'^2 = omega^2 / 2.

Multiple choice horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

The springs of force constants $K, 2K, 4K, 8K,....128K$ are connected vertically in series and a body of mass M is suspended from the last spring. If this system is set into oscillations, the time period will be- 

  1. $2\pi \sqrt{\dfrac{2K}{M}}$
  2. $2\pi \sqrt{\dfrac{M}{K}}$
  3. $2\pi \sqrt{\dfrac{2M}{K}}$
  4. $2\pi \sqrt{\dfrac{M}{2K}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

A spring is placed in vertical position by suspending it from a hook at its top. A similar hook on the bottom of the spring is at $11\ cm$ above a table top. A mass of $75\ g$ and of negligible size is then suspended from the bottom hook, which is measured to be $4.5\ cm$ above the table top. The mass is then pulled down a distance of $4\ cm$ and released. Find the approximate position of the bottom hook after $s$?
Take $g=10m/{s}^{2}$ and hooks mass to be negligible.

  1. $5cm$ above the table top
  2. $4.5cm$ above the table top
  3. $9cm$ above the table top
  4. $0.5cm$ above the table top
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

A block of mass m moving with speed v compresses a spring through distance X before its speed is halved.What is the value of spring constant?

  1. $\frac{{3m{v^2}}}{{4{x^2}}}$
  2. $\frac{{m{v^2}}}{{4{x^2}}}$
  3. $\frac{{m{v^2}}}{{2{x^2}}}$
  4. $\frac{{2m{v^2}}}{{{x^2}}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using work-energy theorem: change in kinetic energy = work done by spring. 1/2 * m * v^2 - 1/2 * m * (v/2)^2 = 1/2 * k * x^2. This simplifies to 1/2 * m * v^2 * (1 - 1/4) = 1/2 * k * x^2, so 3/4 * m * v^2 = k * x^2. Thus, k = 3mv^2 / 4x^2.

Multiple choice horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

A spring executes SHM with mass of 10 kg attached to it. The force constant of spring is 10 N/m.If at any instant its velocity is 40 cm/sec. The displacement will be (here amplitude is0.5m) 

  1. 0.06 m

  2. 0.3 m

  3. 0.01 m

  4. 1.0 m
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In SHM, v = omega * sqrt(A^2 - x^2). Here, omega = sqrt(k/m) = sqrt(10/10) = 1 rad/s. Given v = 0.4 m/s, A = 0.5 m: 0.4 = 1 * sqrt(0.5^2 - x^2). Squaring both sides: 0.16 = 0.25 - x^2, so x^2 = 0.09, x = 0.3 m.

Multiple choice horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

Two identical springs are fixed at one end and masses 1$\mathrm { kg }$ and 4$\mathrm { kg }$ are suspended at their other ends. Theyare both stretched down from their mean position and let go simultaneously. If they are in the same phase atter every 4 seconds then the springs constant $\mathrm { k }$ is 

  1. $\pi \frac { N } { m }$
  2. $\pi ^ { 2 } \frac { \mathrm { N } } { \mathrm { m } }$
  3. 2$\pi \frac { \mathrm { N } } { \mathrm { m } }$

  4. given data is insufficient
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

T1 = 2*pi*sqrt(1/k), T2 = 2*pi*sqrt(4/k) = 4*pi*sqrt(1/k). They are in phase when their time periods are multiples of the common interval. The difference in time periods or their LCM relates to the 4s interval.

Multiple choice horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

A spring mass system is hanging from the ceiling of an elevator in equilibrium. The elevator suddenly starts accelerating upwards with accelerating a, consider all the options in the reference frame of elevator. 

  1. the frequency of oscillation is $\dfrac { 1 }{ 2\pi } \sqrt { \dfrac { k }{ m } } $
  2. the amplitude of the resulting SHM is $\dfrac { ma }{ k } $
  3. amplitude of resulting SHM is $\dfrac { m\left( g+a \right) }{ k } $
  4. maximum speed of block during oscillation is $\left( \sqrt { \dfrac { m }{ k } } \right) a$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Frequency =$2 \pi(m/k)$ frequecy  is independent of $g$ in spring

Extension in spring in equilibrium
initial $ = \dfrac{{mg}}{k}$
Extension in spring in equilibrium  in accelerating lift $m(g+a)$
Amplitude $ = \dfrac{{m\left( {g + a} \right)}}{k} - \dfrac{{mg}}{k} = \dfrac{{ma}}{k}$ 

Multiple choice horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

The reading of a spring balance when a block is suspended from it in air is $60 N$. The reading is changed to $40 N$ when the block is submerged in water. The specific gravity of the block must be therefore.

  1. 3/2

  2. 6

  3. 2

  4. 3
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Apparent weight in air = mg = 60N. Buoyant force = weight in air - weight in water = 60 - 40 = 20N. Buoyant force = density_water * volume * g. Weight = density_block * volume * g. Specific gravity = density_block / density_water = (Weight_air / V*g) / (Buoyant_force / V*g) = 60 / 20 = 3.

Multiple choice horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

A block of mass $1$ kg is connected to a spring of spring constant $\pi^2 N/m$ fixed at other end and kept on smooth level ground. The block is pulled by a distance of $1$ cm from natural length position and released. After what time does the block compress the spring by $\frac{1}{2} cm$.

  1. $\dfrac{2}{3}$ sec
  2. $\dfrac{1}{3}$ sec
  3. $\dfrac{1}{6}$ sec
  4. $\dfrac{1}{12}$ sec
Reveal answer Fill a bubble to check yourself
C Correct answer