
A block of mass m = 1kg is placed on a smooth surface and is connected with a spring of constant K = 100N/m and another end of spring is connected to a fixed wall as shown. The block is pulled by a distance A = 0.10 m from its natural length and released at t = 0
(A) The maximum speed is after $t=\dfrac{\pi }{20}s$
(B) Time taken to cover the first 0.10 m, $t=\dfrac{\pi }{20}s$
(C) Time taken to cover the first 0.05 m, $t=\dfrac{\pi }{40}s$
(D) Time taken to cover the first 0.05 m, $t=\dfrac{\pi }{30}s$
Answer
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Hint: In the question, a smooth floor is given which means that there is no friction in between the block and the floor, so, no amount of energy will be lost due to friction. Also, this case is a case of simple harmonic motion of a spring, which means it will oscillate at a specific frequency for a particular time period.
Formula used:
For solving this question, we will be using the formula for angular frequency of a spring in simple harmonic motion, i.e.,
$\omega =\sqrt{\dfrac{K}{m}}$
And, the formula
$T=\dfrac{2\pi }{\omega }$
Complete answer:
First, let us take a look at the given parameters,
K = 100N/m
m = 1 kg
L = 1 m
Now, using the formula for the angular frequency of a spring in simple harmonic motion,
$\omega =\sqrt{\dfrac{K}{m}}$
Using the given parameters
$\Rightarrow \omega =\sqrt{\dfrac{100}{1}}$
$\Rightarrow \omega =10rad/s$
Now that we have the angular acceleration,
So, using
$T=\dfrac{2\pi }{\omega }$
Now, the maximum speed of the spring will be at its natural length,
So,
$\Rightarrow \dfrac{T}{4}=\dfrac{2\pi }{10\times 4}$
$\Rightarrow \dfrac{T}{4}=\dfrac{\pi }{20}s$
Now, the time taken to cover 0.1m is $\dfrac{T}{4}=\dfrac{\pi }{20}s$
Now, Time taken to cover 0.05 m will be
$\Rightarrow \dfrac{T}{4}\times \dfrac{2}{3}$
$\Rightarrow \dfrac{\pi }{20}\times \dfrac{2}{3}$
$\Rightarrow \dfrac{\pi }{30}s$
So, the correct options will be,
Option (A) The maximum speed is after $t=\dfrac{\pi }{20}s$ , Option (B) Time taken to cover the first 0.10 m, $t=\dfrac{\pi }{20}s$ and Option (D) Time taken to cover the first 0.05 m, $t=\dfrac{\pi }{30}s$
Note:
Simple harmonic motion is a special form of periodic motion in dynamics and physics where the restoring force on the moving object is directly proportional to the displacement magnitude of the object and works against the equilibrium position of the object.
Formula used:
For solving this question, we will be using the formula for angular frequency of a spring in simple harmonic motion, i.e.,
$\omega =\sqrt{\dfrac{K}{m}}$
And, the formula
$T=\dfrac{2\pi }{\omega }$
Complete answer:
First, let us take a look at the given parameters,
K = 100N/m
m = 1 kg
L = 1 m
Now, using the formula for the angular frequency of a spring in simple harmonic motion,
$\omega =\sqrt{\dfrac{K}{m}}$
Using the given parameters
$\Rightarrow \omega =\sqrt{\dfrac{100}{1}}$
$\Rightarrow \omega =10rad/s$
Now that we have the angular acceleration,
So, using
$T=\dfrac{2\pi }{\omega }$
Now, the maximum speed of the spring will be at its natural length,
So,
$\Rightarrow \dfrac{T}{4}=\dfrac{2\pi }{10\times 4}$
$\Rightarrow \dfrac{T}{4}=\dfrac{\pi }{20}s$
Now, the time taken to cover 0.1m is $\dfrac{T}{4}=\dfrac{\pi }{20}s$
Now, Time taken to cover 0.05 m will be
$\Rightarrow \dfrac{T}{4}\times \dfrac{2}{3}$
$\Rightarrow \dfrac{\pi }{20}\times \dfrac{2}{3}$
$\Rightarrow \dfrac{\pi }{30}s$
So, the correct options will be,
Option (A) The maximum speed is after $t=\dfrac{\pi }{20}s$ , Option (B) Time taken to cover the first 0.10 m, $t=\dfrac{\pi }{20}s$ and Option (D) Time taken to cover the first 0.05 m, $t=\dfrac{\pi }{30}s$
Note:
Simple harmonic motion is a special form of periodic motion in dynamics and physics where the restoring force on the moving object is directly proportional to the displacement magnitude of the object and works against the equilibrium position of the object.
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